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

Q.1

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

Ans

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

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

$\begin{array}{l}\text{Factorise\hspace{0.33em}the\hspace{0.33em}following\hspace{0.33em}using\hspace{0.33em}appropriate\hspace{0.33em}identities:}\\ \left(\text{i}\right){\text{\hspace{0.33em}9x}}^{\text{2}}{\text{+6xy+y}}^{\text{2}}\\ \left(\text{ii}\right){\text{\hspace{0.33em}4y}}^{\text{2}}-\text{4y+1}\\ \left(\text{iii}\right){\text{\hspace{0.33em}x}}^{\text{2}}-\frac{{\text{y}}^{\text{2}}}{\text{100}}\end{array}$

Ans

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

Q.4

$\begin{array}{l}\text{Expand\hspace{0.33em}each\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}following,\hspace{0.33em}using\hspace{0.33em}suitable\hspace{0.33em}identities:}\\ \left(\text{i}\right){\left(\text{x+2y+4z}\right)}^{\text{2}}\\ \left(\text{ii}\right){\left(\text{2x}-\text{y+z}\right)}^{\text{2}}\\ \left(\text{iii}\right){\left(-\text{2x+3y+2z}\right)}^{\text{2}}\\ \left(\text{iv}\right){\left(\text{3a}-\text{7b}-\text{c}\right)}^{\text{2}}\\ \left(\mathrm{vi}\right){\left(-\text{2x+5y}-\text{3z}\right)}^{\text{2}}\\ \left(\mathrm{vii}\right){\left[\frac{\text{1}}{\text{4}}\text{a}-\frac{\text{1}}{\text{2}}\text{b+1}\right]}^{\text{2}}\end{array}$

Ans

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

Q.5

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

Ans

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

Q.6

$\begin{array}{l}\text{Write\hspace{0.33em}the\hspace{0.33em}following\hspace{0.33em}cubes in\hspace{0.33em}expanded\hspace{0.33em}form:}\\ \left(\text{i}\right){\left(\text{2x+1}\right)}^{\text{3}}\\ \left(\text{ii}\right){\left(\text{2a-3b}\right)}^{\text{3}}\\ \left(\text{iii}\right){\left[\frac{\text{3}}{\text{2}}\text{x+1}\right]}^{\text{3}}\\ \left(\text{iv}\right){\left[\text{x-}\frac{\text{2}}{\text{3}}\text{y}\right]}^{\text{3}}\end{array}$

Ans

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

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

Ans

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

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{8a}}^{\text{3}}+{\text{b}}^{\text{3}}+\text{12}{a}^{\text{2}}b+\text{6}a{\text{b}}^{\text{2}}\\ ={\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}}\\ ={\left(2a+b\right)}^3\\ \left[\because {\left(a+b\right)}^3=a^3+3a^{2}b+3a{b}^2+b^3\right]\\ =\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{8}{a}^{\text{3}}–{\text{b}}^{\text{3}}–{\text{12a}}^{\text{2}}\text{b}+{\text{6ab}}^{\text{2}}\\ ={\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}}\\ ={\left(2a-b\right)}^3\\ \left[\because {\left(a-b\right)}^3=a^3-3a^{2}b+3a{b}^2-b^3\right]\\ =\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}}\\ ={\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}}\\ ={\left(3-5a\right)}^3\\ \left[\because {\left(a-b\right)}^3=a^3-3a^{2}b+3a{b}^2-b^3\right]\\ =\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}}\\ ={\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}}\\ ={\left(4a-3b\right)}^3\\ \left[\because {\left(a-b\right)}^3=a^3-3a^{2}b+3a{b}^2-b^3\right]\\ =\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}}–\left(\frac{\text{1}}{216}\right)–\left(\frac{\text{9}}{2}\right){\text{p}}^{\text{2}}+\left(\frac{\text{1}}{4}\right)\text{p}\\ ={\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}}\\ ={\left(3p-\frac{1}{6}\right)}^3\\ \left[\because {\left(a-b\right)}^3=a^3-3a^{2}b+3a{b}^2-b^3\right]\\ =\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) (x² – xy+ 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) (x² – xy+ 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) (x² – xy+ 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) (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)
(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

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

Ans

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

$$ $\mathrm{OR}$

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

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

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

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

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

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

Q.15

$\begin{array}{l}\text{Give\hspace{0.33em}possible\hspace{0.33em}expressions\hspace{0.33em}for\hspace{0.33em}the\hspace{0.33em}length\hspace{0.33em}and\hspace{0.33em}breadth\hspace{0.33em}of}\\ \text{each\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}following\hspace{0.33em}rectangles,\hspace{0.33em}in\hspace{0.33em}which\hspace{0.33em}their\hspace{0.33em}areas are}\\ \text{given:}\\ \left(\text{i}\right){\text{\hspace{0.33em}Area\hspace{0.33em}:\hspace{0.33em}25a}}^{\text{2}}\text{-35a+12}\\ \left(\text{ii}\right){\text{\hspace{0.33em}Area\hspace{0.33em}:\hspace{0.33em}35y}}^{\text{2}}\text{+13y-12}\end{array}$