# NCERT Solutions for class 9 Maths Chapter 2- Polynomials Exercise 2.5

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## NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.5 (Ex 2.5)

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## Access NCERT Solutions for Maths Chapter 2 – Polynomials

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### Exercise 2.5

Class 9 Maths Chapter 2 Exercise 2.5 is based on the topic of Algebraic Identities. There are eight identities that students need to focus on. These identities can be used to factorise the polynomials.

### Summary of Polynomial Chapter

The chapter on polynomials is a crucial chapter for students to practice. Students should study this chapter thoroughly and practice the exercises. They can refer to Extramarks for solutions. Extramarks also provides revision notes that can help in summarising the topics.

### Degree of a Polynomial

The coefficients of the variables in a polynomial are called their degree. This topic builds the foundation for the concepts of Polynomials.

### Types of Polynomial

There are various types of polynomials that students may encounter. These include linear polynomials, quadratic polynomials, etc.

### Value of a Polynomial

Another important concept is the Value of a Polynomial. Students should study the NCERT book thoroughly to fully understand the concept.

### Zero of a Polynomial

The zero of a polynomial is a real number that can be found by equating the polynomial to zero. This real number is also called the root of that equation.

### Remainder Theorem

The remainder theorem is one of the most important ways of factorising a polynomial. It is one of the most crucial topics in the chapter. Students should solve the NCERT exercises to be thorough with the chapter.

### Factor Theorem

Another important method of factorising a polynomial is the Factor Theorem. Students should practice numerous questions based on this topic as it is very important for the exams.

### Algebraic Identities

There are various algebraic identities that are useful for mastering the concepts of Polynomials. Students should solve Exercise 2.5 Class 9 Math entirely to be thorough. There are 16 questions in Class 9 Maths Ex 2.5, that are very helpful to prepare for exams.

### Benefits of NCERT Solutions Provided by Extramarks

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### NCERT Solutions for Class 9

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### CBSE Study Materials for Class 9

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### CBSE Study Materials

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Q.1

$\begin{array}{l}\text{Use suitable identities to find the following products:}\\ \left(\text{i}\right)\text{ }\left(\text{x+4}\right)\left(\text{x+10}\right)\\ \left(\text{ii}\right)\text{ }\left(\text{x+8}\right)\left(\text{x}-\text{10}\right)\\ \left(\text{iii}\right)\text{ }\left(\text{3x+4}\right)\left(\text{3x}-\text{5}\right)\\ \left(\text{iv}\right)\text{ }\left({\text{y}}^{\text{2}}\text{+}\frac{\text{3}}{\text{2}}\right)\left({\text{y}}^{\text{2}}-\frac{\text{3}}{\text{2}}\right)\\ \left(\text{v}\right)\left(\text{3}-\text{2x}\right)\left(\text{3+2x}\right)\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{x}+4\right)\left(\mathrm{x}+10\right)={\mathrm{x}}^{2}+4\mathrm{x}+10\mathrm{x}+40\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{2}+14\mathrm{x}+40\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{x}+8\right)\left(\mathrm{x}-10\right)={\mathrm{x}}^{2}-10\mathrm{x}+8\mathrm{x}-80\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{2}-2\mathrm{x}-80\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(3\mathrm{x}+4\right)\left(3\mathrm{x}-5\right)=9{\mathrm{x}}^{2}-15\mathrm{x}+12\mathrm{x}-20\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=9{\mathrm{x}}^{2}-3\mathrm{x}-20\\ \left(\mathrm{iv}\right)\left({\mathrm{y}}^{2}+\frac{3}{2}\right)\left({\mathrm{y}}^{2}-\frac{3}{2}\right)={\left({\mathrm{y}}^{2}\right)}^{2}-{\left(\frac{3}{2}\right)}^{2}\left[\because \text{\hspace{0.17em}}\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}\right)={\mathrm{a}}^{2}-{\mathrm{b}}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{y}}^{4}-\frac{9}{4}\\ \left(\mathrm{v}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(3-2\mathrm{x}\right)\left(3+2\mathrm{x}\right)={\left(3\right)}^{2}-{\left(2\mathrm{x}\right)}^{2}\left[\because \text{\hspace{0.17em}}\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}\right)={\mathrm{a}}^{2}-{\mathrm{b}}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=9-4{\mathrm{x}}^{2}\end{array}$

Q.2 Evaluate the following products without multiplying directly: (i) 103 × 107 (ii) 95 × 96 (iii) 104 × 9

Ans
(i) 103 × 107 = (100 + 3)(100 + 7)
= 10000 + (3 + 7)100 + 21
= 10000 + 1000 + 21
= 11021 (ii) 95 × 96 = (100 – 5)( 100 – 4)
= 10000 – (5 + 4)100 + 20
= 10000 – 900 + 20
= 9120 (iii) 104 × 96 = (100 + 4)(100 – 4)
= 10000 – (4)2 [(a – b)(a + b) = a2 – b2 ]
= 10000 – 16
= 9984

Q.3

$\begin{array}{l}\text{Factorise​ the following using appropriate identities:}\\ \left(\text{i}\right){\text{ 9x}}^{\text{2}}{\text{+6xy+y}}^{\text{2}}\\ \left(\text{ii}\right){\text{ 4y}}^{\text{2}}-\text{4y+1}\\ \left(\text{iii}\right){\text{ x}}^{\text{2}}-\frac{{\text{y}}^{\text{2}}}{\text{100}}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{ }9{\mathrm{x}}^{2}+6\mathrm{xy}+{\mathrm{y}}^{2}={\left(3\mathrm{x}\right)}^{2}+2\left(3\mathrm{x}\right)\mathrm{y}+{\mathrm{y}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left(3\mathrm{x}+\mathrm{y}\right)}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\because {\mathrm{a}}^{2}+2\mathrm{ab}+{\mathrm{b}}^{2}={\left(\mathrm{a}+\mathrm{b}\right)}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(3\mathrm{x}+\mathrm{y}\right)\left(3\mathrm{x}+\mathrm{y}\right)\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4{\mathrm{y}}^{2}-4\mathrm{y}+1={\left(2\mathrm{y}\right)}^{2}-2\left(2\mathrm{y}\right)1+{\left(1\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(2\mathrm{y}-1\right)\left(2\mathrm{y}-1\right)\left[\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}\right)={\mathrm{a}}^{2}-{\mathrm{b}}^{2}\right]\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-\frac{{\mathrm{y}}^{2}}{100}={\mathrm{x}}^{2}-{\left(\frac{\mathrm{y}}{10}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{x}-\frac{\mathrm{y}}{10}\right)\left(\mathrm{x}+\frac{\mathrm{y}}{10}\right)\left[\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}\right)={\mathrm{a}}^{2}-{\mathrm{b}}^{2}\right]\end{array}$

Q.4

$\begin{array}{l}\text{Expand each of the following, using suitable identities:}\\ \left(\text{i}\right)\text{ }{\left(\text{x+2y+4z}\right)}^{\text{2}}\\ \left(\text{ii}\right)\text{ }{\left(\text{2x}-\text{y+z}\right)}^{\text{2}}\\ \left(\text{iii}\right)\text{ }{\left(-\text{2x+3y+2z}\right)}^{\text{2}}\\ \left(\text{iv}\right)\text{ }{\left(\text{3a}-\text{7b}-\text{c}\right)}^{\text{2}}\\ \left(\mathrm{vi}\right)\text{ }{\left(-\text{2x+5y}-\text{3z}\right)}^{\text{2}}\\ \left(\mathrm{vii}\right)\text{ }{\left[\frac{\text{1}}{\text{4}}\text{a}-\frac{\text{1}}{\text{2}}\text{b+1}\right]}^{\text{2}}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{ }{\left(\mathrm{x}+2\mathrm{y}+4\mathrm{z}\right)}^{2}={\mathrm{x}}^{2}+{\left(2\mathrm{y}\right)}^{2}+{\left(4\mathrm{z}\right)}^{2}+2\mathrm{x}\left(2\mathrm{y}\right)+2\left(2\mathrm{y}\right)\left(4\mathrm{z}\right)+2\left(4\mathrm{z}\right)\mathrm{x}\\ \text{\hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{2}+4{\mathrm{y}}^{2}+16{\mathrm{z}}^{2}+4\mathrm{xy}+16\mathrm{yz}+8\mathrm{zx}\\ \left[\because {\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+{\mathrm{c}}^{2}+2\mathrm{ab}+2\mathrm{bc}+2\mathrm{ca}\right]\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}}{\left(2\mathrm{x}-\mathrm{y}+\mathrm{z}\right)}^{2}={\left(2\mathrm{x}\right)}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}-2\left(2\mathrm{x}\right)\mathrm{y}-2\mathrm{yz}+2\mathrm{z}\left(2\mathrm{x}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=4{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}-4\mathrm{xy}-2\mathrm{yz}+4\mathrm{zx}\\ \left[\because {\left(\mathrm{a}-\mathrm{b}+\mathrm{c}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+{\mathrm{c}}^{2}-2\mathrm{ab}-2\mathrm{bc}+2\mathrm{ca}\right]\\ \left(\mathrm{iii}\right)\text{ }{\left(-2\mathrm{x}+3\mathrm{y}+2\mathrm{z}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left\{-\left(2\mathrm{x}-3\mathrm{y}-2\mathrm{z}\right)\right\}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left(2\mathrm{x}-3\mathrm{y}-2\mathrm{z}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left(2\mathrm{x}\right)}^{2}+{\left(3\mathrm{y}\right)}^{2}+{\left(2\mathrm{z}\right)}^{2}-2\left(2\mathrm{x}\right)\left(3\mathrm{y}\right)+2\left(3\mathrm{y}\right)\left(2\mathrm{z}\right)-2\left(2\mathrm{z}\right)\left(2\mathrm{x}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4{\mathrm{x}}^{2}+9{\mathrm{y}}^{2}+4{\mathrm{z}}^{2}-12\mathrm{xy}+12\mathrm{yz}-8\mathrm{zx}\\ \left[\because {\left(\mathrm{a}-\mathrm{b}-\mathrm{c}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+{\mathrm{c}}^{2}-2\mathrm{ab}+2\mathrm{bc}-2\mathrm{ca}\right]\\ \left(\mathrm{iv}\right)\text{ }{\left(3\mathrm{a}-7\mathrm{b}-\mathrm{c}\right)}^{2}\text{\hspace{0.17em}\hspace{0.17em}}={\left(3\mathrm{a}\right)}^{2}+{\left(7\mathrm{b}\right)}^{2}+{\left(\mathrm{c}\right)}^{2}-2\left(3\mathrm{a}\right)\left(7\mathrm{b}\right)+2\left(7\mathrm{b}\right)\mathrm{c}-2\mathrm{c}\left(3\mathrm{a}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=9{\mathrm{a}}^{2}+49{\mathrm{b}}^{2}+{\mathrm{c}}^{2}-42\mathrm{ab}+14\mathrm{bc}-6\mathrm{ca}\\ \left[\because {\left(\mathrm{a}-\mathrm{b}-\mathrm{c}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+{\mathrm{c}}^{2}-2\mathrm{ab}+2\mathrm{bc}-2\mathrm{ca}\right]\end{array}$ $\begin{array}{l}\left(\mathrm{v}\right){\left(-2\mathrm{x}+5\mathrm{y}-3\mathrm{z}\right)}^{2}\\ ={\left\{-\left(2\mathrm{x}-5\mathrm{y}+3\mathrm{z}\right)\right\}}^{2}\\ ={\left(2\mathrm{x}-5\mathrm{y}+3\mathrm{z}\right)}^{2}\\ ={\left(2\mathrm{x}\right)}^{2}+25{\mathrm{y}}^{2}+9{\mathrm{z}}^{2}-2\left(2\mathrm{x}\right)\left(5\mathrm{y}\right)-2\left(5\mathrm{y}\right)\left(3\mathrm{z}\right)+2\left(3\mathrm{z}\right)\left(2\mathrm{x}\right)\\ =4{\mathrm{x}}^{2}+25{\mathrm{y}}^{2}+9{\mathrm{z}}^{2}-20\mathrm{xy}-30\mathrm{yz}+12\mathrm{zx}\\ \left[\because {\left(\mathrm{a}-\mathrm{b}+\mathrm{c}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+{\mathrm{c}}^{2}-2\mathrm{ab}-2\mathrm{bc}+2\mathrm{ca}\right]\\ \left(\mathrm{vi}\right)\text{\hspace{0.17em}}{\left[\frac{1}{4}\mathrm{a}-\frac{1}{2}\mathrm{b}+1\right]}^{2}\\ ={\left(\frac{1}{4}\mathrm{a}\right)}^{2}+{\left(\frac{1}{2}\mathrm{b}\right)}^{2}+1-2\left(\frac{1}{4}\mathrm{a}\right)\left(\frac{1}{2}\mathrm{b}\right)-2\left(\frac{1}{2}\mathrm{b}\right)\left(1\right)+2\left(1\right)\left(\frac{1}{4}\mathrm{a}\right)\\ =\frac{1}{16}{\mathrm{a}}^{2}+\frac{1}{4}{\mathrm{b}}^{2}+1-\frac{1}{4}\mathrm{ab}-\mathrm{b}+\frac{1}{2}\mathrm{a}\\ \left[\because {\left(\mathrm{a}-\mathrm{b}+\mathrm{c}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+{\mathrm{c}}^{2}-2\mathrm{ab}-2\mathrm{bc}+2\mathrm{ca}\right]\end{array}$

Q.5

$\begin{array}{l}\mathrm{Factorise}:\\ \left(\mathrm{i}\right)\text{ }4{\mathrm{x}}^{2}+9{\mathrm{y}}^{2}+16{\mathrm{z}}^{2}+12\mathrm{xy}-24\mathrm{yz}-16\mathrm{xz}\\ \left(\mathrm{ii}\right)\text{ }2{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+8{\mathrm{z}}^{2}-2\sqrt{2}\mathrm{xy}+4\sqrt{2}\mathrm{yz}-8\mathrm{xz}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)4{\mathrm{x}}^{2}+9{\mathrm{y}}^{2}+16{\mathrm{z}}^{2}+12\mathrm{xy}-24\mathrm{yz}-16\mathrm{xz}\\ ={\left(2\mathrm{x}\right)}^{2}+{\left(3\mathrm{y}\right)}^{2}+{\left(4\mathrm{z}\right)}^{2}+2\left(2\mathrm{x}\right)\left(3\mathrm{y}\right)-2\left(3\mathrm{y}\right)\left(4\mathrm{z}\right)-2\left(4\mathrm{z}\right)\left(2\mathrm{x}\right)\\ ={\left(2\mathrm{x}+3\mathrm{y}-4\mathrm{z}\right)}^{2}\left[\because {\left(\mathrm{x}+\mathrm{y}-\mathrm{z}\right)}^{2}={\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}+2\mathrm{xy}-2\mathrm{yz}-2\mathrm{zx}\right]\\ =\left(2\mathrm{x}+3\mathrm{y}-4\mathrm{z}\right)\left(2\mathrm{x}+3\mathrm{y}-4\mathrm{z}\right)\\ \\ \left(\mathrm{ii}\right)2{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+8{\mathrm{z}}^{2}-2\sqrt{2}\text{\hspace{0.17em}}\mathrm{xy}+4\sqrt{2}\mathrm{yz}-8\mathrm{xz}\\ ={\left(\sqrt{2}\mathrm{x}\right)}^{2}+{\left(\mathrm{y}\right)}^{2}+{\left(2\sqrt{2}\mathrm{z}\right)}^{2}-2\left(\sqrt{2}\mathrm{x}\right)\left(\mathrm{y}\right)+2\left(\mathrm{y}\right)\left(2\sqrt{2}\mathrm{z}\right)-2\left(2\sqrt{2}\mathrm{z}\right)\left(\sqrt{2}\mathrm{x}\right)\\ =\left(\sqrt{2}\mathrm{x}-\mathrm{y}-2\sqrt{2}\mathrm{z}\right)\left(\sqrt{2}\mathrm{x}-\mathrm{y}-2\sqrt{2}\mathrm{z}\right)\\ \left[\because {\left(\mathrm{x}-\mathrm{y}-\mathrm{z}\right)}^{2}={\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}-2\mathrm{xy}+2\mathrm{yz}-2\mathrm{zx}\right]\end{array}$

Q.6

$\begin{array}{l}\text{Write​ the following cubes in expanded form:}\\ \left(\text{i}\right)\text{ }{\left(\text{2x+1}\right)}^{\text{3}}\\ \left(\text{ii}\right)\text{ }{\left(\text{2a-3b}\right)}^{\text{3}}\\ \left(\text{iii}\right)\text{ }{\left[\frac{\text{3}}{\text{2}}\text{x+1}\right]}^{\text{3}}\\ \left(\text{iv}\right)\text{ }{\left[\text{x-}\frac{\text{2}}{\text{3}}\text{y}\right]}^{\text{3}}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(2\mathrm{x}+1\right)}^{3}={\left(2\mathrm{x}\right)}^{3}+3\left(2\mathrm{x}\right)1\left(2\mathrm{x}+1\right)+{\left(1\right)}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8{\mathrm{x}}^{3}+6\mathrm{x}\left(2\mathrm{x}+1\right)+1\\ \left[\because {\left(\mathrm{a}+\mathrm{b}\right)}^{3}={\mathrm{a}}^{3}+3\mathrm{ab}\left(\mathrm{a}+\mathrm{b}\right)+{\mathrm{b}}^{3}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8{\mathrm{x}}^{3}+12{\mathrm{x}}^{2}+6\mathrm{x}+1\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}}{\left(2\mathrm{a}-3\mathrm{b}\right)}^{3}={\left(2\mathrm{a}\right)}^{3}-3\left(2\mathrm{a}\right)\left(3\mathrm{b}\right)\left(2\mathrm{a}-3\mathrm{b}\right)-{\left(3\mathrm{b}\right)}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8{\mathrm{a}}^{3}-18\mathrm{ab}\left(2\mathrm{a}-3\mathrm{b}\right)-27{\mathrm{b}}^{3}\\ \left[\because {\left(\mathrm{a}-\mathrm{b}\right)}^{3}={\mathrm{a}}^{3}-3\mathrm{ab}\left(\mathrm{a}-\mathrm{b}\right)-{\mathrm{b}}^{3}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8{\mathrm{a}}^{3}-36{\mathrm{a}}^{2}\mathrm{b}+54{\mathrm{ab}}^{2}-27{\mathrm{b}}^{3}\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left[\frac{3}{2}\mathrm{x}+1\right]}^{3}={\left(\frac{3}{2}\mathrm{x}\right)}^{3}+3\left(\frac{3}{2}\mathrm{x}\right)1\left(\frac{3}{2}\mathrm{x}+1\right)+{1}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{27}{8}{\mathrm{x}}^{3}+\frac{9}{2}\mathrm{x}\left(\frac{3}{2}\mathrm{x}+1\right)+1\\ \left[\because {\left(\mathrm{a}+\mathrm{b}\right)}^{3}={\mathrm{a}}^{3}+3\mathrm{ab}\left(\mathrm{a}+\mathrm{b}\right)+{\mathrm{b}}^{3}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{27}{8}{\mathrm{x}}^{3}+\frac{27}{4}{\mathrm{x}}^{2}+\frac{9}{2}\mathrm{x}+1\\ \left(\mathrm{iv}\right)\text{ }{\left[\mathrm{x}-\frac{2}{3}\mathrm{y}\right]}^{3}={\left(\mathrm{x}\right)}^{3}-3\left(\mathrm{x}\right)\left(\frac{2}{3}\mathrm{y}\right)\left(\mathrm{x}-\frac{2}{3}\mathrm{y}\right)-{\left(\frac{2}{3}\mathrm{y}\right)}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{3}-2{\mathrm{x}}^{2}\mathrm{y}+\frac{4}{9}{\mathrm{xy}}^{2}-\frac{8}{27}{\mathrm{y}}^{3}\end{array}$

Q.7 Evaluate the following using suitable identities: (i) (99)3 (ii) (102)3 (iii) (998)3

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(99\right)}^{3}={\left(100–1\right)}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left(100\right)}^{3}-3\left(100\right).1\left(100-1\right)-{1}^{3}\text{\hspace{0.17em}}\left[\because {\left(\mathrm{a}-\mathrm{b}\right)}^{3}={\mathrm{a}}^{3}-3\mathrm{ab}\left(\mathrm{a}-\mathrm{b}\right)-{\mathrm{b}}^{3}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1000000-300×99-1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1000000-29700-1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=970299\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(102\right)}^{3}={\left(100+2\right)}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{}{\left(100\right)}^{3}+3\left(100\right)\left(2\right)\left(100+2\right)+{2}^{3}\left[\because {\left(\mathrm{a}+\mathrm{b}\right)}^{3}={\mathrm{a}}^{3}+3\mathrm{ab}\left(\mathrm{a}+\mathrm{b}\right)+{\mathrm{b}}^{3}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1000000+600×102+8\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1000000+61200+8\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1061208\\ \left(\mathrm{iii}\right)\text{}{\left(998\right)}^{3}={\left(1000–2\right)}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left(1000\right)}^{3}-3\left(1000\right)\left(2\right)\left(1000-2\right)-{2}^{3}\left[\because {\left(\mathrm{a}-\mathrm{b}\right)}^{3}={\mathrm{a}}^{3}-3\mathrm{ab}\left(\mathrm{a}-\mathrm{b}\right)-{\mathrm{b}}^{3}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1000000000-6000×998-8\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=994011992\end{array}$

Q.8 Factorise each of the following:
(i) 8a3 + b3 + 12a2b + 6ab2
(ii) 8a3 – b3 – 12a2b + 6ab2
(iii) 27 – 125a3 – 135a + 225a2
(iv) 64a3 – 27b3 – 144a2b + 108ab2
(v) 27p3 – (1/216) –(9/2)p2 + (1/4)p

Ans

$\begin{array}{l}\left(\text{i}\right)\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{8a}}^{\text{3}}+{\text{b}}^{\text{3}}+\text{12}{a}^{\text{2}}b+\text{6}a{\text{b}}^{\text{2}}\\ \text{}\text{}\text{}\text{}\text{}={\left(\text{2a}\right)}^{\text{3}}+{\text{b}}^{\text{3}}+\text{3}{\left(2a\right)}^{\text{2}}b+\text{3}\left(2a\right){\text{b}}^{\text{2}}\\ \text{}\text{}\text{}\text{}\text{}={\left(2a+b\right)}^{3}\text{\hspace{0.17em}}\\ \text{}\text{}\text{}\text{}\text{}\text{}\left[\because {\left(a+b\right)}^{3}={a}^{3}+3{a}^{2}b+3a{b}^{2}+{b}^{3}\right]\\ \text{}\text{}\text{}\text{}\text{}=\left(2a+b\right)\left(2a+b\right)\left(2a+b\right)\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\left(\text{ii}\right)\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{8}{a}^{\text{3}}–{\text{b}}^{\text{3}}–{\text{12a}}^{\text{2}}\text{b}+{\text{6ab}}^{\text{2}}\\ \text{}\text{}\text{}\text{}\text{}={\left(\text{2a}\right)}^{\text{3}}-{\text{b}}^{\text{3}}-\text{3}{\left(2a\right)}^{\text{2}}b+\text{3}\left(2a\right){\text{b}}^{\text{2}}\\ \text{}\text{}\text{}\text{}\text{}={\left(2a-b\right)}^{3}\text{\hspace{0.17em}}\\ \text{}\text{}\text{}\text{}\text{}\text{}\left[\because {\left(a-b\right)}^{3}={a}^{3}-3{a}^{2}b+3a{b}^{2}-{b}^{3}\right]\\ \text{}\text{}\text{}\text{}\text{}=\left(2a-b\right)\left(2a-b\right)\left(2a-b\right)\\ \left(\text{iii}\right)\text{27}–{\text{125a}}^{\text{3}}–\text{135}a+{\text{225a}}^{\text{2}}\\ \text{}\text{}\text{}\text{}\text{}={\left(\text{3}\right)}^{\text{3}}-{\left(5a\right)}^{\text{3}}-\text{3}{\left(3\right)}^{\text{2}}\left(5a\right)+\text{3}\left(3\right){\left(5a\right)}^{\text{2}}\\ \text{}\text{}\text{}\text{}\text{}={\left(3-5a\right)}^{3}\text{\hspace{0.17em}}\\ \text{}\text{}\text{}\text{}\text{}\text{}\left[\because {\left(a-b\right)}^{3}={a}^{3}-3{a}^{2}b+3a{b}^{2}-{b}^{3}\right]\\ \text{}\text{}\text{}\text{}\text{}=\left(3-5a\right)\left(3-5a\right)\left(3-5a\right)\\ \left(\text{iv}\right)\text{64}{a}^{\text{3}}–{\text{27b}}^{\text{3}}–{\text{144a}}^{\text{2}}\text{b}+\text{1}0{\text{8ab}}^{\text{2}}\\ \text{}\text{}\text{}\text{}\text{}={\left(\text{4a}\right)}^{\text{3}}-{\left(\text{3b}\right)}^{\text{3}}-\text{3}{\left(4a\right)}^{\text{2}}\left(3b\right)+\text{3}\left(4a\right){\left(\text{3b}\right)}^{\text{2}}\\ \text{}\text{}\text{}\text{}\text{}={\left(4a-3b\right)}^{3}\text{\hspace{0.17em}}\\ \text{}\text{}\text{}\text{}\text{}\text{}\left[\because {\left(a-b\right)}^{3}={a}^{3}-3{a}^{2}b+3a{b}^{2}-{b}^{3}\right]\\ \text{}\text{}\text{}\text{}\text{}=\left(4a-3b\right)\left(4a-3b\right)\left(4a-3b\right)\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\left(\text{v}\right){\text{27p}}^{\text{3}}–\text{}\left(\frac{\text{1}}{216}\right)\text{}–\left(\frac{\text{9}}{2}\right){\text{p}}^{\text{2}}+\text{}\left(\frac{\text{1}}{4}\right)\text{p}\\ \text{}\text{}\text{}\text{}\text{}={\left(\text{3p}\right)}^{\text{3}}-{\left(\frac{1}{6}\right)}^{\text{3}}-\text{3}{\left(3p\right)}^{\text{2}}\left(\frac{1}{6}\right)+\text{3}\left(3p\right){\left(\frac{1}{6}\right)}^{\text{2}}\\ \text{}\text{}\text{}\text{}\text{}={\left(3p-\frac{1}{6}\right)}^{3}\text{\hspace{0.17em}}\\ \text{}\text{}\text{}\text{}\text{}\text{}\left[\because {\left(a-b\right)}^{3}={a}^{3}-3{a}^{2}b+3a{b}^{2}-{b}^{3}\right]\\ \text{}\text{}\text{}\text{}\text{}=\left(3p-\frac{1}{6}\right)\left(3p-\frac{1}{6}\right)\left(3p-\frac{1}{6}\right)\end{array}$

Q.9 Verify:
(i) x3+ y3 = (x + y) (x2xy+ y2)
(ii) x3 y3 = (x – y) (x2 + xy + y2)

Ans
(i) x3+ y3 = (x + y) (x2 – xy+ y2)
Putting x = 1, y = 1 in L.H.S., we get
L.H.S. = x3+ y3 = 13+ 13 = 2
R.H.S. = (x + y) (x2xy+ y2)
= (1 + 1) (12 1.1+ 12)
= 2(1 – 1 + 1)
= 2
So, L.H.S. = R.H.S.
Thus, given equation is verified for x = 1 and y = 1.
Again, putting x = 2 and y = 3, we get
L.H.S. = x3+ y3 = 23+ 33 = 35
R.H.S. = (x + y) (x2xy+ y2)
= (2 + 3) (22 – (2).(3)+ 32)
= 5(4 – 6 + 9)
= 35
So, L.H.S. = R.H.S.
Thus, given equation is verified for x = 2 and y = 3.
Again, putting x = 3 and y = 0, we get
L.H.S. = x3+ y3 = 33+ 03 = 27
R.H.S. = (x + y) (x2xy+ y2)
= (3 + 0) (32 – (3).(0)+ 02)
= 3(9 – 0 + 0)
= 27
So, L.H.S. = R.H.S.
Thus, given equation is verified for x = 3 and y = 0.
In this way, it is verified that
x3+ y3 = (x + y) (x2xy+ y2)
(ii) x3 y3 = (x – y) (x2 + xy + y2)
Putting x = 1, y = 2 in L.H.S., we get
L.H.S. = x3 – y3 = 13 – 13 = 0
R.H.S. = (x – y) (x2 + xy + y2)
= (1 1) (12 1.1+ 12)
= 0
So, L.H.S. = R.H.S.
Thus, given equation is verified for x = 1 and y = 1.
Again, putting x = 2 and y = 3, we get
L.H.S. = x3 y3 = 23 33 = 19
R.H.S. = (x y) (x2 + xy+ y2)
= (2 3) (22 + (2).(3)+ 32)
= 1(4 + 6 + 9)
= 19
So, L.H.S. = R.H.S.
Thus, given equation is verified for x = 2 and y = 3.
Again, putting x = 3 and y = 0, we get
L.H.S. = x3 y3 = 33 03 = 27
R.H.S. = (x – y) (x2 + xy+ y2)
= (3 + 0) (32 + (3).(0)+ 02)
= 3(9 + 0 + 0)= 27
So, L.H.S. = R.H.S.
Thus, given equation is verified for x = 3 and y = 0.
In this way, it is verified that
x3 – y3 = (x – y) (x2 + xy + y2)

Q.10 Factorise each of the following:
(i) 27y3+ 125z3
(ii) 64m3 – 343n3

Ans
(i) 27y3+ 125z3= (3y)3 + (5z)3
= (3y + 5z){(3y)2 – (3y)(5z) +(5z)2}
= (3y + 5z)(9y2 – 15yz + 25z2)
[Since, a3 + b3 =(a + b)(a2 – ab + b2)] (ii) 64m3 –343n3= (4m)3 + (7n)3
= (4m – 7n){(4m)2 + (4m)(7n) +(7n)2}
= (4m – 7n)(16m2 + 28mn + 49n2)
[Since, a3 – b3 =(a – b)(a2 + ab + b2)]

Q.11 Factorise: 27x3 + y3 + z3 – 9xyz

Ans
Since,
a3 + b3 + c3 – 3abc
= (a + b +c)(a2 + b2 + c2 – ab – bc – ca)
So, 27x3 + y3 + z3 – 9xyz
= (3x)3 + y3 + z3 – 3(3x)yz
= (3x + y + z){(3x)2 + y2 + z2 –(3x)y – yz – z(3x)}
= (3x + y + z)(9x2 + y2 + z2 – 3xy – yz – 3xz)

Q.12

$\begin{array}{l}\mathrm{Verify}\mathrm{that}:\\ {\mathrm{x}}^{3}+{\mathrm{y}}^{3}+{\mathrm{z}}^{3}–3\mathrm{xyz}=\frac{1}{2}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left[{\left(\mathrm{x}-\mathrm{y}\right)}^{2}+{\left(\mathrm{y}-\mathrm{z}\right)}^{2}+{\left(\mathrm{z}-\mathrm{x}\right)}^{2}\right]\end{array}$

Ans

$\begin{array}{l}\mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{1}{2}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left[{\left(\mathrm{x}-\mathrm{y}\right)}^{2}+{\left(\mathrm{y}-\mathrm{z}\right)}^{2}+{\left(\mathrm{z}-\mathrm{x}\right)}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left[{\mathrm{x}}^{2}-2\mathrm{xy}+{\mathrm{y}}^{2}+{\mathrm{y}}^{2}-2\mathrm{yz}+{\mathrm{z}}^{2}+{\mathrm{z}}^{2}-2\mathrm{zx}+{\mathrm{x}}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left(2{\mathrm{x}}^{2}+2{\mathrm{y}}^{2}+2{\mathrm{z}}^{2}-2\mathrm{xy}-2\mathrm{yz}-2\mathrm{zx}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left({\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}-\mathrm{xy}-\mathrm{yz}-\mathrm{zx}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{3}+{\mathrm{y}}^{3}+{\mathrm{z}}^{3}–3\mathrm{xyz}=\mathrm{L}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Hence}\text{proved.}\end{array}$

 $\mathrm{OR}$

$\begin{array}{l}{\mathrm{x}}^{3}+{\mathrm{y}}^{3}+{\mathrm{z}}^{3}–3\mathrm{xyz}=\frac{1}{2}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left[{\left(\mathrm{x}-\mathrm{y}\right)}^{2}+{\left(\mathrm{y}-\mathrm{z}\right)}^{2}+{\left(\mathrm{z}-\mathrm{x}\right)}^{2}\right]\\ \mathrm{For}\text{verifying the above equation, we have to put different}\\ \text{values of x, y and z in both sides.}\\ \mathrm{Putting}\text{x}=0,\mathrm{y}=1\text{and z}=\text{2 in L.H.S., we get}\\ \text{L.H.S.}={\mathrm{x}}^{3}+{\mathrm{y}}^{3}+{\mathrm{z}}^{3}–3\mathrm{xyz}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={0}^{3}+{1}^{3}+{2}^{3}–3\left(0\right)\left(1\right)\left(2\right)\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}}=1+8\text{\hspace{0.17em}}=9\\ \mathrm{Putting}\text{x}=0,\mathrm{y}=1\text{and z}=\text{2 in R.H.S., we get}\\ \text{R.H.S.}=\frac{1}{2}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left[{\left(\mathrm{x}-\mathrm{y}\right)}^{2}+{\left(\mathrm{y}-\mathrm{z}\right)}^{2}+{\left(\mathrm{z}-\mathrm{x}\right)}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left(0+1+2\right)\left[{\left(0-1\right)}^{2}+{\left(1-2\right)}^{2}+{\left(2-0\right)}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3}{2}×6\text{\hspace{0.17em}}=9\\ \mathrm{Thus},\text{L.H.S.}=\text{R.H.S.}\\ \mathrm{Hence},\text{it is verified that given equation is true for x}=0,\mathrm{y}=1\text{}\\ \text{and z}=\text{2}.\\ \mathrm{Putting}\text{x}=2,\mathrm{y}=5\text{and z}=\text{3 in L.H.S., we get}\\ \text{L.H.S.}={\mathrm{x}}^{3}+{\mathrm{y}}^{3}+{\mathrm{z}}^{3}–3\mathrm{xyz}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={2}^{3}+{5}^{3}+{3}^{3}–3\left(2\right)\left(5\right)\left(3\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8+125+27-90=70\\ \mathrm{Putting}\text{x}=2,\mathrm{y}=5\text{and z}=\text{3 in R.H.S., we get}\\ \text{R.H.S.}=\frac{1}{2}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left[{\left(\mathrm{x}-\mathrm{y}\right)}^{2}+{\left(\mathrm{y}-\mathrm{z}\right)}^{2}+{\left(\mathrm{z}-\mathrm{x}\right)}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left(2+5+3\right)\left[{\left(2-5\right)}^{2}+{\left(5-3\right)}^{2}+{\left(3-2\right)}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}×10×14=70\\ \mathrm{Hence},\text{it is verified that given equation is true for x}=2,\mathrm{y}=5\\ \text{and z}=\text{3}.\\ \mathrm{Putting}\text{x}=11,\mathrm{y}=8\text{and z}=\text{5 in L.H.S., we get}\\ \text{L.H.S.}={\mathrm{x}}^{3}+{\mathrm{y}}^{3}+{\mathrm{z}}^{3}–3\mathrm{xyz}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left(11\right)}^{3}+{8}^{3}+{5}^{3}–3\left(11\right)\left(8\right)\left(5\right)\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}\end{array}$

$\begin{array}{l}=1331+512+125-1320\text{\hspace{0.17em}}=648\\ \mathrm{Putting}\text{x}=11,\mathrm{y}=8\text{and z}=\text{5 in R.H.S., we get}\\ \text{R.H.S.}=\frac{1}{2}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left[{\left(\mathrm{x}-\mathrm{y}\right)}^{2}+{\left(\mathrm{y}-\mathrm{z}\right)}^{2}+{\left(\mathrm{z}-\mathrm{x}\right)}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left(11+8+5\right)\left[{\left(11-8\right)}^{2}+{\left(8-5\right)}^{2}+{\left(5-11\right)}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}×24×54\text{\hspace{0.17em}}=648\\ \mathrm{Thus},\text{L.H.S.}=\text{R.H.S.}\\ \mathrm{Hence},\text{it is verified that given equation is true for x}=11,\mathrm{y}=8\text{}\\ \text{and z}=\text{5}.\\ \mathrm{Thus},\mathrm{it}\text{is verified that}\\ {\mathrm{x}}^{3}+{\mathrm{y}}^{3}+{\mathrm{z}}^{3}–3\mathrm{xyz}=\frac{1}{2}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left[{\left(\mathrm{x}-\mathrm{y}\right)}^{2}+{\left(\mathrm{y}-\mathrm{z}\right)}^{2}+{\left(\mathrm{z}-\mathrm{x}\right)}^{2}\right]\end{array}$

Q.13 If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.

Ans
Since,
a3 + b3 + c3 – 3abc
= (a + b +c)(a2 + b2 + c2 – ab – bc – ca)
So, x3 + y3 + z3 – 3xyz
= (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
Putting x + y + z = 0, we get
x3 + y3 + z3 – 3xyz
= (0)(x2 + y2 + z2 – xy – yz – zx)
x3 + y3 + z3 – 3xyz = 0 x3 + y3 + z3 = 3xyz Hence Proved.

Q.14 Without actually calculating the cubes, find the value of each of the following:
(i) (–12)3 + (7)3 + (5)3
(ii) (28)3 + (–15)3 + (–13)3

Ans

$\begin{array}{l}\left(\text{i}\right)\text{}{\left(–\text{12}\right)}^{\text{3}}+\text{}{\left(\text{7}\right)}^{\text{3}}+\text{}{\left(\text{5}\right)}^{\text{3}}\mathrm{}\\ \mathrm{Let},\text{x}=-12,\text{y}=7\text{and z}=\text{5}\\ \text{then,}\mathrm{x}+\mathrm{y}+\mathrm{z}=-12+7+5\text{\hspace{0.17em}}=0\\ \mathrm{Since},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{3}+\text{}{\mathrm{y}}^{3}+\text{}{\mathrm{z}}^{3}–\text{}3\mathrm{xyz}\mathrm{}\text{\hspace{0.17em}\hspace{0.17em}}=\text{}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left({\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}–\mathrm{xy}–\mathrm{yz}–\mathrm{zx}\right)\\ \text{Putting \hspace{0.17em}x}+\text{y}+\text{z}=\text{}0,\text{we get}\mathrm{}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}}^{\text{3}}+{\text{y}}^{\text{3}}+{\text{z}}^{\text{3}}–\text{3xyz}\mathrm{}\text{\hspace{0.17em}\hspace{0.17em}}=\text{}\left(0\right)\left({\text{x}}^{\text{2}}+{\text{y}}^{\text{2}}+{\text{z}}^{\text{2}}–\text{xy}–\text{yz}–\text{zx}\right)\mathrm{}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}}^{\text{3}}+{\text{y}}^{\text{3}}+{\text{z}}^{\text{3}}–\text{3xyz}=\text{}0\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}}^{\text{3}}+{\text{y}}^{\text{3}}+{\text{z}}^{\text{3}}=\text{3xyz}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}}{\left(–\text{12}\right)}^{\text{3}}+\text{}{\left(\text{7}\right)}^{\text{3}}+\text{}{\left(\text{5}\right)}^{\text{3}}=3\left(-12\right)\left(7\right)\left(5\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-1260\end{array}$

$\begin{array}{l}\left(\text{ii}\right)\text{}{\left(\text{28}\right)}^{\text{3}}+\text{}{\left(–\text{15}\right)}^{\text{3}}+\text{}{\left(–\text{13}\right)}^{\text{3}}\\ \mathrm{}\mathrm{Let},\text{x}=28,\text{y}=-15\text{and z}=-\text{13}\\ \text{then,}\mathrm{x}+\mathrm{y}+\mathrm{z}=28-15-13\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=0\\ \mathrm{Since},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{3}+\text{}{\mathrm{y}}^{3}+\text{}{\mathrm{z}}^{3}–\text{}3\mathrm{xyz}\mathrm{}\text{\hspace{0.17em}\hspace{0.17em}}=\text{}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left({\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}–\mathrm{xy}–\mathrm{yz}–\mathrm{zx}\right)\\ \text{Putting \hspace{0.17em}x}+\text{y}+\text{z}=\text{}0,\text{we get}\mathrm{}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}}^{\text{3}}+{\text{y}}^{\text{3}}+{\text{z}}^{\text{3}}–\text{3xyz}\mathrm{}\text{\hspace{0.17em}\hspace{0.17em}}=\text{}\left(0\right)\left({\text{x}}^{\text{2}}+{\text{y}}^{\text{2}}+{\text{z}}^{\text{2}}–\text{xy}–\text{yz}–\text{zx}\right)\mathrm{}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}}^{\text{3}}+{\text{y}}^{\text{3}}+{\text{z}}^{\text{3}}–\text{3xyz}=\text{}0\mathrm{}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}}^{\text{3}}+{\text{y}}^{\text{3}}+{\text{z}}^{\text{3}}=\text{3xyz}\\ \mathrm{}\therefore {\left(\text{28}\right)}^{\text{3}}+\text{}{\left(–\text{15}\right)}^{\text{3}}+\text{}{\left(–\text{13}\right)}^{\text{3}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\left(28\right)\left(-15\right)\left(-13\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=16380\end{array}$

Q.15

$\begin{array}{l}\text{Give possible expressions for the length and breadth of}\\ \text{each of the following rectangles, in which their areas are}\\ \text{given:}\\ \left(\text{i}\right){\text{ Area : 25a}}^{\text{2}}\text{-35a+12}\\ \left(\text{ii}\right){\text{ Area : 35y}}^{\text{2}}\text{+13y-12}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{ }\mathrm{Area}\text{ }:\text{ }25{\mathrm{a}}^{2}-35\mathrm{a}+12=25{\mathrm{a}}^{2}-35\mathrm{a}+12\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=25{\mathrm{a}}^{2}-20\mathrm{a}-15\mathrm{a}+12\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5\mathrm{a}\left(5\mathrm{a}-4\right)-3\left(5\mathrm{a}-4\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(5\mathrm{a}-4\right)\left(5\mathrm{a}-3\right)\\ \mathrm{So},\text{the length of rectangle is}\left(\text{5a}-4\right)\text{and breadth is\hspace{0.17em}}\left(5\mathrm{a}-3\right).\\ \left(\mathrm{ii}\right)\text{ }\mathrm{Area}\text{ }:\text{ }35{\mathrm{y}}^{2}+13\mathrm{y}-12=35{\mathrm{y}}^{2}+28\mathrm{y}-15\mathrm{y}-12\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=7\mathrm{y}\left(5\mathrm{y}+4\right)-3\left(5\mathrm{y}+4\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(5\mathrm{y}+4\right)\left(7\mathrm{y}-3\right)\\ \mathrm{So},\text{the length of rectangle is}\left(7\mathrm{y}-3\right)\text{units and breadth is}\\ \left(5\mathrm{y}+4\right)\text{ }\mathrm{units}.\end{array}$

Q.16 What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume : 3x2–12x
(ii) Volume: 12ky2 + 8ky – 20k

Ans

$\begin{array}{c}\left(\mathrm{i}\right)\mathrm{Volume}=3{\mathrm{x}}^{2}-12\mathrm{x}\\ =3\mathrm{x}\left(\mathrm{x}-4\right)\\ \mathrm{Thus},\text{\hspace{0.17em}\hspace{0.17em}the possible length of cuboid}=3\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{units}\\ \text{breadth of cuboid}=\mathrm{x}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{units}\\ \text{\hspace{0.17em}height of cuboid}=\left(\mathrm{x}-4\right)\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{units}\\ \left(\mathrm{ii}\right)\mathrm{Volume}={\text{12ky}}^{\text{2}}+\text{8ky}-\text{2}0\text{k}\\ =4\mathrm{k}\left(3{\mathrm{y}}^{2}+2\mathrm{y}-5\right)\\ =4\mathrm{k}\left(3{\mathrm{y}}^{2}+5\mathrm{y}-3\mathrm{y}-5\right)\\ =4\mathrm{k}\left(3\mathrm{y}+5\right)\left(\mathrm{y}-1\right)\\ \mathrm{Thus},\text{\hspace{0.17em}\hspace{0.17em}the}\mathrm{possible}\text{length of cuboid}=3\mathrm{k}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{units}\\ \text{breadth of cuboid}=\left(3\mathrm{y}+5\right)\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{units}\\ \text{height of cuboid}=\left(\mathrm{y}-1\right)\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{units}\end{array}$

## FAQs (Frequently Asked Questions)

### 1. Where can students studying in Class 9 obtain the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.5?

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### 2. How should students study for exams using the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.5?

Students should first read the entire chapter, from beginning to end. It is important to learn all of the equation-solving formulas and techniques. They should learn all the methods to factorise the polynomials.  The examples from the NCERT textbook should be solved next. Students will learn several approaches for answering the chapter’s questions as a result. They should try the exercises’ questions after finishing the examples. Without referring to the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.5, they must attempt to solve the problems. As soon as they are finished, they can double-check their answers by referring to the Class 9 Maths Exercise 2.5 Solutions. If they want to correctly respond to the questions, they must fix their mistakes and learn the techniques described in the solutions. They should then retry solving the exercises’ questions to gauge their improvement. Since they will be more equipped to understand the chapter’s material and resolve a variety of equations, they will score better on exams.