NCERT Solutions for Class 9 Maths Ganit Manjari Chapter 4 Exploring Algebraic Identities

Algebraic identities are equations that are true for all values of the variables. In Class 9 Maths Chapter 4, students learn how identities help expand expressions, factorise algebraic expressions, simplify rational expressions and calculate products or squares quickly.

NCERT Solutions for Class 9 Maths Ganit Manjari Chapter 4 help students understand Exploring Algebraic Identities, where algebraic expressions are expanded, factorised and simplified using standard identities. The chapter focuses on square identities, cube identities, factorisation and rational algebraic expressions.

The chapter uses geometrical models, squares, rectangles, cubes, algebra tiles and factorisation methods to show how identities work. These exercise-wise NCERT Solutions for Class 9 Maths Chapter 4 are useful for homework, classroom assignments, CBSE 2026-27 school exam preparation and concept revision.

NCERT Solutions for Class 9 Maths Chapter 4 PDF Download

The NCERT Solutions for Class 9 Maths Chapter 4 PDF gives students a formula-focused and solution-focused way to revise algebraic identities. It is helpful for practising expansions, factorisation, cube identities and simplification questions that commonly appear in school exams.

Students can use the PDF to revise:

1. Algebraic identities
2. Difference between equation and identity
3. Expansion using identities
4. Factorisation using identities
5. Algebra tiles
6. Products of binomials
7. Factorisation of quadratic expressions
8. Cube identities
9. Sum and difference of cubes
10. Simplification of rational algebraic expressions

Access Exercise-Wise NCERT Solutions for Class 9 Maths Chapter 4

Exercise	Main Focus	Solution Type
Exercise Set 4.1	Expansion using (a + b)²	Identity-based expansion and calculations
Exercise Set 4.2	Factorisation using square identities	Perfect square factorisation
Exercise Set 4.3	More identities	Squares, factorisation and three-term identities
Exercise Set 4.4	Factorisation without algebra tiles	Quadratic factorisation and products
Exercise Set 4.5	Rational algebraic expressions	Simplification using factorisation
End-of-Chapter Exercises	Mixed identity revision	Expansion, factorisation, simplification and word problems

NCERT Solutions for Class 9 Maths Chapter 4 Exercise Set 4.1

Exercise Set 4.1 focuses on expanding expressions using the identity (a + b)² = a² + 2ab + b² and using the same identity for quick numerical calculations.

Question 1. Using the identity (a + b)² = a² + 2ab + b², expand the following.

(i) (7x + 4y)²

Answer:
Here:

a = 7x, b = 4y

(7x + 4y)² = (7x)² + 2(7x)(4y) + (4y)²
= 49x² + 56xy + 16y²

Final answer:
49x² + 56xy + 16y²

(ii) (7x/5 + 3y/2)²

Answer:
Here:

a = 7x/5, b = 3y/2

(7x/5 + 3y/2)²
= (7x/5)² + 2(7x/5)(3y/2) + (3y/2)²
= 49x²/25 + 21xy/5 + 9y²/4

Final answer:
49x²/25 + 21xy/5 + 9y²/4

(iii) (2.5p + 1.5q)²

Answer:
Here:

a = 2.5p, b = 1.5q

(2.5p + 1.5q)²
= (2.5p)² + 2(2.5p)(1.5q) + (1.5q)²
= 6.25p² + 7.5pq + 2.25q²

Final answer:
6.25p² + 7.5pq + 2.25q²

(iv) (3s/4 + 8t)²

Answer:
Here:

a = 3s/4, b = 8t

(3s/4 + 8t)²
= (3s/4)² + 2(3s/4)(8t) + (8t)²
= 9s²/16 + 12st + 64t²

Final answer:
9s²/16 + 12st + 64t²

(v) (x + 1/y)²

Answer:
Here:

a = x, b = 1/y

(x + 1/y)²
= x² + 2(x)(1/y) + (1/y)²
= x² + 2x/y + 1/y²

Final answer:
x² + 2x/y + 1/y²

(vi) (1/x + 1/y)²

Answer:
Here:

a = 1/x, b = 1/y

(1/x + 1/y)²
= (1/x)² + 2(1/x)(1/y) + (1/y)²
= 1/x² + 2/xy + 1/y²

Final answer:
1/x² + 2/xy + 1/y²

Question 2. Using the same identity, find the values of the following.

(i) 64²

Answer:
64 = 60 + 4

64² = (60 + 4)²
= 60² + 2(60)(4) + 4²
= 3600 + 480 + 16
= 4096

Final answer:
4096

(ii) 105²

Answer:
105 = 100 + 5

105² = (100 + 5)²
= 100² + 2(100)(5) + 5²
= 10000 + 1000 + 25
= 11025

Final answer:
11025

(iii) 205²

Answer:
205 = 200 + 5

205² = (200 + 5)²
= 200² + 2(200)(5) + 5²
= 40000 + 2000 + 25
= 42025

Final answer:
42025

NCERT Solutions for Class 9 Maths Chapter 4 Exercise Set 4.2

Exercise Set 4.2 focuses on factorisation using the identities (a + b)² = a² + 2ab + b² and (a - b)² = a² - 2ab + b².

Question 1. Factor completely.

(i) 9x² + 24xy + 16y²

Answer:
9x² = (3x)²
16y² = (4y)²
24xy = 2(3x)(4y)

So:

9x² + 24xy + 16y² = (3x + 4y)²

Final answer:
(3x + 4y)²

(ii) 4s² + 20st + 25t²

Answer:
4s² = (2s)²
25t² = (5t)²
20st = 2(2s)(5t)

So:

4s² + 20st + 25t² = (2s + 5t)²

Final answer:
(2s + 5t)²

(iii) 49x² + 28xy + 4y²

Answer:
49x² = (7x)²
4y² = (2y)²
28xy = 2(7x)(2y)

So:

49x² + 28xy + 4y² = (7x + 2y)²

Final answer:
(7x + 2y)²

(iv) 64p² + 32pq/3 + 4q²/9

Answer:
64p² = (8p)²
4q²/9 = (2q/3)²
32pq/3 = 2(8p)(2q/3)

So:

64p² + 32pq/3 + 4q²/9 = (8p + 2q/3)²

Final answer:
(8p + 2q/3)²

(v) 3a² + 4ab + 4b²/3

Answer:
Take 1/3 common:

3a² + 4ab + 4b²/3
= 1/3(9a² + 12ab + 4b²)

Now:

9a² = (3a)²
4b² = (2b)²
12ab = 2(3a)(2b)

So:

1/3(9a² + 12ab + 4b²)
= 1/3(3a + 2b)²

Final answer:
(3a + 2b)²/3

(vi) 9s²/5 + 6sv + 5v²

Answer:
Take 1/5 common:

9s²/5 + 6sv + 5v²
= 1/5(9s² + 30sv + 25v²)

Now:

9s² = (3s)²
25v² = (5v)²
30sv = 2(3s)(5v)

So:

1/5(9s² + 30sv + 25v²)
= 1/5(3s + 5v)²

Final answer:
(3s + 5v)²/5

Question 2. Find the values using the identity (a - b)² = a² - 2ab + b².

(i) 79²

Answer:
79 = 80 - 1

79² = (80 - 1)²
= 80² - 2(80)(1) + 1²
= 6400 - 160 + 1
= 6241

Final answer:
6241

(ii) 193²

Answer:
193 = 200 - 7

193² = (200 - 7)²
= 200² - 2(200)(7) + 7²
= 40000 - 2800 + 49
= 37249

Final answer:
37249

(iii) 299²

Answer:
299 = 300 - 1

299² = (300 - 1)²
= 300² - 2(300)(1) + 1²
= 90000 - 600 + 1
= 89401

Final answer:
89401

NCERT Solutions for Class 9 Maths Chapter 4 Exercise Set 4.3

Exercise Set 4.3 covers the identities (a + b)², (a - b)² and (a + b + c)². It also includes factorisation using suitable identities.

Question 1. Find the following squares using one of the identities.

(i) 117²

Answer:
117 = 100 + 10 + 7

117² = (100 + 10 + 7)²
= 100² + 10² + 7² + 2(100)(10) + 2(10)(7) + 2(7)(100)
= 10000 + 100 + 49 + 2000 + 140 + 1400
= 13689

Final answer:
13689

(ii) 78²

Answer:
78 = 80 - 2

78² = (80 - 2)²
= 80² - 2(80)(2) + 2²
= 6400 - 320 + 4
= 6084

Final answer:
6084

(iii) 198²

Answer:
198 = 200 - 2

198² = (200 - 2)²
= 200² - 2(200)(2) + 2²
= 40000 - 800 + 4
= 39204

Final answer:
39204

(iv) 214²

Answer:
214 = 200 + 10 + 4

214² = (200 + 10 + 4)²
= 200² + 10² + 4² + 2(200)(10) + 2(10)(4) + 2(4)(200)
= 40000 + 100 + 16 + 4000 + 80 + 1600
= 45796

Final answer:
45796

(v) 1104²

Answer:
1104 = 1100 + 4

1104² = (1100 + 4)²
= 1100² + 2(1100)(4) + 4²
= 1210000 + 8800 + 16
= 1218816

Final answer:
1218816

(vi) 1120²

Answer:
1120 = 1100 + 20

1120² = (1100 + 20)²
= 1100² + 2(1100)(20) + 20²
= 1210000 + 44000 + 400
= 1254400

Final answer:
1254400

Question 2. Factor using suitable identities.

(i) 16y² - 24y + 9

Answer:
16y² = (4y)²
9 = 3²
24y = 2(4y)(3)

So:

16y² - 24y + 9 = (4y - 3)²

Final answer:
(4y - 3)²

(ii) 9s²/4 + 6st + 4t²

Answer:
9s²/4 = (3s/2)²
4t² = (2t)²
6st = 2(3s/2)(2t)

So:

9s²/4 + 6st + 4t² = (3s/2 + 2t)²

Final answer:
(3s/2 + 2t)²

(iii) m² + 9k² + 4n² + 6mk + 4mn + 12kn

Answer:
Compare with:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Take:

a = m, b = 3k, c = 2n

Then:

m² + 9k² + 4n² + 6mk + 12kn + 4mn = (m + 3k + 2n)²

Final answer:
(m + 3k + 2n)²

(iv) p² - 2p/16 + 1/16²

Answer:
This expression matches:

a² - 2ab + b²

Take:

a = p, b = 1/16

So:

p² - 2p/16 + 1/16² = (p - 1/16)²

Final answer:
(p - 1/16)²

(v) 9a² + 4b² + c² + 12ab - 6ac - 4bc

Answer:
Compare with:

(x + y - z)² = x² + y² + z² + 2xy - 2xz - 2yz

Take:

x = 3a, y = 2b, z = c

Then:

9a² + 4b² + c² + 12ab - 6ac - 4bc
= (3a + 2b - c)²

Final answer:
(3a + 2b - c)²

Question 3. Expand using the identity (a + b + c)².

(i) (p + 3q + 7r)²

Answer:
Here:

a = p, b = 3q, c = 7r

(p + 3q + 7r)²
= p² + 9q² + 49r² + 2(p)(3q) + 2(3q)(7r) + 2(7r)(p)
= p² + 9q² + 49r² + 6pq + 42qr + 14pr

Final answer:
p² + 9q² + 49r² + 6pq + 42qr + 14pr

(ii) (3x - 2y + 4z)²

Answer:
Here:

a = 3x, b = -2y, c = 4z

(3x - 2y + 4z)²
= (3x)² + (-2y)² + (4z)² + 2(3x)(-2y) + 2(-2y)(4z) + 2(4z)(3x)
= 9x² + 4y² + 16z² - 12xy - 16yz + 24xz

Final answer:
9x² + 4y² + 16z² - 12xy - 16yz + 24xz

Question 4. Is this an identity?

(a + b - c)² + (a - b + c)² + (-a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Answer:
Expand the left-hand side.

(a + b - c)² = a² + b² + c² + 2ab - 2ac - 2bc

(a - b + c)² = a² + b² + c² - 2ab + 2ac - 2bc

(-a + b + c)² = a² + b² + c² - 2ab - 2ac + 2bc

Adding:

LHS = 3a² + 3b² + 3c² - 2ab - 2ac - 2bc

This is not equal to:

a² + b² + c² + 2ab + 2bc + 2ca

for all values of a, b and c.

Final answer:
No, it is not an identity.

NCERT Solutions for Class 9 Maths Chapter 4 Exercise Set 4.4

Exercise Set 4.4 focuses on factorisation without algebra tiles. Students learn how to split the middle term and factor quadratic expressions.

Question 1. Fill in the blanks to complete the following identities.

(i) s² - 11s + 24 = ()()

Answer:
We need two numbers whose sum is -11 and product is 24.

The numbers are -3 and -8.

s² - 11s + 24 = (s - 3)(s - 8)

Final answer:
(s - 3)(s - 8)

(ii) (________)(x + 1) = 3x² - 4x - 7

Answer:
Factor 3x² - 4x - 7.

We need two numbers whose product is 3 × (-7) = -21 and sum is -4.

The numbers are -7 and 3.

3x² - 4x - 7
= 3x² - 7x + 3x - 7
= x(3x - 7) + 1(3x - 7)
= (3x - 7)(x + 1)

Final answer:
3x - 7

(iii) 10x² - 11x - 6 = (2x - )( + 2)

Answer:
Factor 10x² - 11x - 6.

10x² - 11x - 6
= 10x² - 15x + 4x - 6
= 5x(2x - 3) + 2(2x - 3)
= (2x - 3)(5x + 2)

Final answer:
10x² - 11x - 6 = (2x - 3)(5x + 2)

(iv) 6x² + 7x + 2 = (________)(_______)

Answer:
We need two numbers whose product is 6 × 2 = 12 and sum is 7.

The numbers are 3 and 4.

6x² + 7x + 2
= 6x² + 3x + 4x + 2
= 3x(2x + 1) + 2(2x + 1)
= (3x + 2)(2x + 1)

Final answer:
(3x + 2)(2x + 1)

Question 2. Select and use the identity that will help you find the products without multiplying directly.

(i) 41²

Answer:
41 = 40 + 1

41² = (40 + 1)²
= 1600 + 80 + 1
= 1681

Final answer:
1681

(ii) 27²

Answer:
27 = 30 - 3

27² = (30 - 3)²
= 900 - 180 + 9
= 729

Final answer:
729

(iii) 23 × 17

Answer:
23 × 17 = (20 + 3)(20 - 3)

Use:

(a + b)(a - b) = a² - b²

= 20² - 3²
= 400 - 9
= 391

Final answer:
391

(iv) 135²

Answer:
135 = 100 + 30 + 5

135² = (100 + 30 + 5)²
= 10000 + 900 + 25 + 6000 + 300 + 3000
= 20225

Final answer:
20225

(v) 97²

Answer:
97 = 100 - 3

97² = (100 - 3)²
= 10000 - 600 + 9
= 9409

Final answer:
9409

(vi) 18 × 29

Answer:
18 × 29 = 18 × (30 - 1)
= 18 × 30 - 18
= 540 - 18
= 522

Final answer:
522

(vii) 34 × 43

Answer:
34 × 43 = 34 × (40 + 3)
= 34 × 40 + 34 × 3
= 1360 + 102
= 1462

Final answer:
1462

(viii) 205²

Answer:
205 = 200 + 5

205² = (200 + 5)²
= 40000 + 2000 + 25
= 42025

Final answer:
42025

Question 3. Factor the following.

(i) 9a² + b² + 4c² - 6ab + 12ac - 4bc

Answer:
Compare with:

(x - y + z)² = x² + y² + z² - 2xy + 2xz - 2yz

Take:

x = 3a, y = b, z = 2c

So:

9a² + b² + 4c² - 6ab + 12ac - 4bc = (3a - b + 2c)²

Final answer:
(3a - b + 2c)²

(ii) 16s² + 25t² - 40st

Answer:
16s² = (4s)²
25t² = (5t)²
40st = 2(4s)(5t)

So:

16s² + 25t² - 40st = (4s - 5t)²

Final answer:
(4s - 5t)²

(iii) r² - r - 42

Answer:
We need two numbers whose product is -42 and sum is -1.

The numbers are -7 and 6.

r² - r - 42
= r² - 7r + 6r - 42
= r(r - 7) + 6(r - 7)
= (r + 6)(r - 7)

Final answer:
(r + 6)(r - 7)

(iv) 49g² + 14gh + h²

Answer:
49g² = (7g)²
h² = h²
14gh = 2(7g)(h)

So:

49g² + 14gh + h² = (7g + h)²

Final answer:
(7g + h)²

(v) 64u² + 121v² + 4w² - 176uv - 32uw + 44vw

Answer:
Compare with:

(x - y - z)² = x² + y² + z² - 2xy - 2xz + 2yz

Take:

x = 8u, y = 11v, z = 2w

Then:

64u² + 121v² + 4w² - 176uv - 32uw + 44vw
= (8u - 11v - 2w)²

Final answer:
(8u - 11v - 2w)²

NCERT Solutions for Class 9 Maths Chapter 4 Exercise Set 4.5

Exercise Set 4.5 focuses on simplifying rational algebraic expressions by factorising the numerator and denominator and cancelling common factors.

Question 1. Simplify the following rational expressions, assuming denominators are not equal to zero.

(i) (3p² - 3pq - 18q²) / (3p² + 3pq - 10q²)

Answer:
Factor numerator:

3p² - 3pq - 18q²
= 3(p² - pq - 6q²)
= 3(p - 3q)(p + 2q)

Factor denominator:

3p² + 3pq - 10q²
= (3p - 2q)(p + 5q)

There is no common factor.

Final answer:
3(p - 3q)(p + 2q) / [(3p - 2q)(p + 5q)]

(ii) (n³ - 3n²m + 3nm² - m³) / (5m² - 10mn + 5n²)

Answer:
Numerator:

n³ - 3n²m + 3nm² - m³ = (n - m)³

Denominator:

5m² - 10mn + 5n²
= 5(m² - 2mn + n²)
= 5(m - n)²
= 5(n - m)²

So:

(n - m)³ / 5(n - m)² = (n - m)/5

Final answer:
(n - m)/5

(iii) (w³ - v³ + x³ + 3wvx) / (w² + v² + x² - 2wv - 2vx + 2wx)

Answer:
Rewrite numerator:

w³ + x³ - v³ + 3wvx

This can be written as:

w³ + x³ + (-v)³ - 3(w)(x)(-v)

Using:

a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

Take:

a = w, b = x, c = -v

Numerator = (w + x - v)(w² + x² + v² - wx + xv + vw)

The denominator:

w² + v² + x² - 2wv - 2vx + 2wx

This equals:

(w + x - v)²

So the simplified expression is:

[w² + x² + v² - wx + xv + vw] / (w + x - v)

Final answer:
(w² + v² + x² - wx + xv + vw) / (w + x - v)

(iv) (4y² - 20yz + 25z²) / (25z² - 4y²)

Answer:
Numerator:

4y² - 20yz + 25z² = (2y - 5z)²

Denominator:

25z² - 4y² = (5z - 2y)(5z + 2y)

Since:

5z - 2y = -(2y - 5z)

So:

(2y - 5z)² / [(5z - 2y)(5z + 2y)]
= (2y - 5z)² / [-(2y - 5z)(5z + 2y)]
= -(2y - 5z)/(5z + 2y)

Final answer:
(5z - 2y)/(5z + 2y)

(v) [(x² + 6x - 7)(x² - 12x + 12)] / [(x² - 6x + 8)(x² - 9)]

Answer:
Factor each expression:

x² + 6x - 7 = (x + 7)(x - 1)

x² - 6x + 8 = (x - 2)(x - 4)

x² - 9 = (x - 3)(x + 3)

The expression x² - 12x + 12 does not factor into simple integer factors.

Final answer:
[(x + 7)(x - 1)(x² - 12x + 12)] / [(x - 2)(x - 4)(x - 3)(x + 3)]

(vi) (p⁴ - 16) / (p² - 4p + 4)

Answer:
Numerator:

p⁴ - 16 = (p²)² - 4²
= (p² - 4)(p² + 4)
= (p - 2)(p + 2)(p² + 4)

Denominator:

p² - 4p + 4 = (p - 2)²

So:

(p⁴ - 16)/(p² - 4p + 4)
= [(p - 2)(p + 2)(p² + 4)]/(p - 2)²
= [(p + 2)(p² + 4)]/(p - 2)

Final answer:
(p + 2)(p² + 4)/(p - 2)

NCERT Solutions for Class 9 Maths Chapter 4 End-of-Chapter Exercises

The end-of-chapter exercises include questions on identity-based products, numerical calculations, factorisation, rational expressions, dimensions of rectangles and cuboids, reciprocal equations and proof-based algebra questions.

Question 1. Use suitable identities to find the following products.

(i) (-3x + 4)²

Answer:
(-3x + 4)² = (4 - 3x)²
= 4² - 2(4)(3x) + (3x)²
= 16 - 24x + 9x²

Final answer:
9x² - 24x + 16

(ii) (2s + 7)(2s - 7)

Answer:
Use:

(a + b)(a - b) = a² - b²

(2s + 7)(2s - 7)
= (2s)² - 7²
= 4s² - 49

Final answer:
4s² - 49

(iii) (p² + 1/2)(p² - 1/2)

Answer:
Use:

(a + b)(a - b) = a² - b²

Here:

a = p², b = 1/2

(p² + 1/2)(p² - 1/2)
= (p²)² - (1/2)²
= p⁴ - 1/4

Final answer:
p⁴ - 1/4

(iv) (2n + 7)(2n - 7)

Answer:
(2n + 7)(2n - 7)
= (2n)² - 7²
= 4n² - 49

Final answer:
4n² - 49

(v) (s - 2t)(s² + 2st + 4t²)

Answer:
Use:

a³ - b³ = (a - b)(a² + ab + b²)

Here:

a = s, b = 2t

(s - 2t)(s² + 2st + 4t²)
= s³ - (2t)³
= s³ - 8t³

Final answer:
s³ - 8t³

(vi) (1/2r - 4r²)²

Answer:
Use:

(a - b)² = a² - 2ab + b²

Take:

a = 1/(2r), b = 4r²

(1/2r - 4r²)²
= 1/(4r²) - 2(1/2r)(4r²) + 16r⁴
= 1/(4r²) - 4r + 16r⁴

Final answer:
1/(4r²) - 4r + 16r⁴

(vii) (-3m + 4k - l)²

Answer:
Use:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Take:

a = -3m, b = 4k, c = -l

(-3m + 4k - l)²
= 9m² + 16k² + l² - 24mk - 8kl + 6ml

Final answer:
9m² + 16k² + l² - 24mk - 8kl + 6ml

(viii) (x - y/3)³

Answer:
Use:

(a - b)³ = a³ - 3a²b + 3ab² - b³

Take:

a = x, b = y/3

(x - y/3)³
= x³ - 3x²(y/3) + 3x(y²/9) - y³/27
= x³ - x²y + xy²/3 - y³/27

Final answer:
x³ - x²y + xy²/3 - y³/27

(ix) (7k/2 - 2m/3)³

Answer:
Use:

(a - b)³ = a³ - 3a²b + 3ab² - b³

Take:

a = 7k/2, b = 2m/3

a³ = 343k³/8

3a²b = 3 × 49k²/4 × 2m/3 = 49k²m/2

3ab² = 3 × 7k/2 × 4m²/9 = 14km²/3

b³ = 8m³/27

Final answer:
343k³/8 - 49k²m/2 + 14km²/3 - 8m³/27

Question 2. Find the values using suitable identities.

(i) 17 × 21

Answer:
17 × 21 = (19 - 2)(19 + 2)

= 19² - 2²
= 361 - 4
= 357

Final answer:
357

(ii) 104 × 96

Answer:
104 × 96 = (100 + 4)(100 - 4)

= 100² - 4²
= 10000 - 16
= 9984

Final answer:
9984

(iii) 24 × 16

Answer:
24 × 16 = (20 + 4)(20 - 4)

= 20² - 4²
= 400 - 16
= 384

Final answer:
384

(iv) 147³

Answer:
147 = 150 - 3

147³ = (150 - 3)³
= 150³ - 3(150)²(3) + 3(150)(3²) - 3³
= 3375000 - 202500 + 4050 - 27
= 3171523

Final answer:
3171523

(v) 199³

Answer:
199 = 200 - 1

199³ = (200 - 1)³
= 200³ - 3(200)²(1) + 3(200)(1²) - 1³
= 8000000 - 120000 + 600 - 1
= 7880599

Final answer:
7880599

(vi) 127³

Answer:
127 = 120 + 7

127³ = (120 + 7)³
= 120³ + 3(120)²(7) + 3(120)(7²) + 7³
= 1728000 + 302400 + 17640 + 343
= 2048383

Final answer:
2048383

(vii) (-107)³

Answer:
(-107)³ = -(107³)

107 = 100 + 7

107³ = 100³ + 3(100)²(7) + 3(100)(7²) + 7³
= 1000000 + 210000 + 14700 + 343
= 1225043

So:

(-107)³ = -1225043

Final answer:
-1225043

(viii) (-299)³

Answer:
(-299)³ = -(299³)

299 = 300 - 1

299³ = 300³ - 3(300)²(1) + 3(300)(1²) - 1
= 27000000 - 270000 + 900 - 1
= 26730900 - 1
= 26730899

So:

(-299)³ = -26730899

Final answer:
-26730899

Question 3. Factor the following algebraic expressions.

(i) 4y² + 1/y² + 1/16

Answer:
This expression appears to require a middle term to form a perfect square. As written in the uploaded text, it is incomplete for direct factorisation by standard identities.

Final answer:
The expression needs to be checked in the textbook image or PDF before final factorisation.

(ii) 9m² - n²/25

Answer:
Use:

a² - b² = (a + b)(a - b)

9m² - n²/25
= (3m)² - (n/5)²
= (3m + n/5)(3m - n/5)

Final answer:
(3m + n/5)(3m - n/5)

(iii) 27b³ - 1/(64b³)

Answer:
Use:

a³ - b³ = (a - b)(a² + ab + b²)

27b³ = (3b)³
1/(64b³) = (1/4b)³

So:

27b³ - 1/(64b³)
= (3b - 1/4b)(9b² + 3/4 + 1/16b²)

Final answer:
(3b - 1/4b)(9b² + 3/4 + 1/16b²)

(iv) x² + 5x/6 + 1/6

Answer:
We need two numbers whose product is 1/6 and sum is 5/6.

The numbers are 1/2 and 1/3.

x² + 5x/6 + 1/6
= x² + x/2 + x/3 + 1/6
= x(x + 1/2) + 1/3(x + 1/2)
= (x + 1/2)(x + 1/3)

Final answer:
(x + 1/2)(x + 1/3)

(v) 27u³/125 - 27u²/25 + 9u/5 - 1

Answer:
This matches:

(a - b)³ = a³ - 3a²b + 3ab² - b³

Take:

a = 3u/5, b = 1

So:

27u³/125 - 27u²/25 + 9u/5 - 1
= (3u/5 - 1)³

Final answer:
(3u/5 - 1)³

(vi) 64y³ + z³/125

Answer:
Use:

a³ + b³ = (a + b)(a² - ab + b²)

64y³ = (4y)³
z³/125 = (z/5)³

So:

64y³ + z³/125
= (4y + z/5)(16y² - 4yz/5 + z²/25)

Final answer:
(4y + z/5)(16y² - 4yz/5 + z²/25)

(vii) p³ + q³ + 27r³ - 9pqr

Answer:
Use:

x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)

Take:

x = p, y = q, z = 3r

Then:

p³ + q³ + 27r³ - 9pqr
= (p + q + 3r)(p² + q² + 9r² - pq - 3qr - 3pr)

Final answer:
(p + q + 3r)(p² + q² + 9r² - pq - 3qr - 3pr)

(viii) 9m² - 12m + 4

Answer:
9m² = (3m)²
4 = 2²
12m = 2(3m)(2)

So:

9m² - 12m + 4 = (3m - 2)²

Final answer:
(3m - 2)²

(ix) 9x³ - 8y³ + 6x²y - 12xy²

Answer:
Rearrange:

9x³ + 6x²y - 12xy² - 8y³

This does not directly match a standard identity as written. It may require checking the original textbook formatting because the uploaded text appears to have formatting distortion.

Final answer:
The expression should be verified from the textbook image/PDF before final factorisation.

(x) 4x² + 9y² + 36z² + 12xy + 36yz + 24xz

Answer:
Compare with:

(a + b + c)²

Take:

a = 2x, b = 3y, c = 6z

Then:

4x² + 9y² + 36z² + 12xy + 36yz + 24xz
= (2x + 3y + 6z)²

Final answer:
(2x + 3y + 6z)²

(xi) 27u³ - 1/216 - 9u²/2 + u/4

Answer:
This matches:

(a - b)³

Take:

a = 3u, b = 1/6

(3u - 1/6)³
= 27u³ - 3(9u²)(1/6) + 3(3u)(1/36) - 1/216
= 27u³ - 9u²/2 + u/4 - 1/216

Final answer:
(3u - 1/6)³

Question 4. Simplify the following, assuming denominators are not equal to 0.

(i) (4x² + 4x + 1)/(4x² - 1)

Answer:
Numerator:

4x² + 4x + 1 = (2x + 1)²

Denominator:

4x² - 1 = (2x + 1)(2x - 1)

So:

(4x² + 4x + 1)/(4x² - 1)
= (2x + 1)/(2x - 1)

Final answer:
(2x + 1)/(2x - 1)

(ii) [9(a + b)³ - 3(a + b)] / [9a² - 36b²]

Answer:
The uploaded text appears to show this expression with formatting distortion. Interpreting it as:

[9(a + b)³ - 3(a + b)] / [9a² - 36b²]

Numerator:

3(a + b)[3(a + b)² - 1]

Denominator:

9(a² - 4b²)
= 9(a - 2b)(a + 2b)

Final answer:
3(a + b)[3(a + b)² - 1] / [9(a - 2b)(a + 2b)]

This question should be checked against the textbook image/PDF for exact formatting before final upload.

(iii) (s³ + 125t³)/(s² - 2st - 35t²)

Answer:
Numerator:

s³ + 125t³ = s³ + (5t)³
= (s + 5t)(s² - 5st + 25t²)

Denominator:

s² - 2st - 35t²
= (s - 7t)(s + 5t)

Cancel common factor s + 5t.

Final answer:
(s² - 5st + 25t²)/(s - 7t)

Question 5. Find possible expressions for the length and breadth of rectangles whose areas are given.

(i) 25a² - 30ab + 9b²

Answer:
25a² = (5a)²
9b² = (3b)²
30ab = 2(5a)(3b)

So:

25a² - 30ab + 9b² = (5a - 3b)²

Final answer:
Length = 5a - 3b, breadth = 5a - 3b

(ii) 36s² - 49t²

Answer:
Use:

a² - b² = (a + b)(a - b)

36s² - 49t²
= (6s)² - (7t)²
= (6s + 7t)(6s - 7t)

Final answer:
Length = 6s + 7t, breadth = 6s - 7t

Question 6. Find possible expressions for the length, breadth and height of cuboids whose volumes are given.

(i) 6a² - 24b²

Answer:
Take 6 common:

6a² - 24b² = 6(a² - 4b²)
= 6(a + 2b)(a - 2b)

Final answer:
Possible dimensions are 6, a + 2b and a - 2b.

(ii) 3ps² - 15ps + 12p

Answer:
Take 3p common:

3ps² - 15ps + 12p
= 3p(s² - 5s + 4)

Now factor:

s² - 5s + 4 = (s - 1)(s - 4)

Final answer:
Possible dimensions are 3p, s - 1 and s - 4.

Question 7. The village playground is shaped as a square of side 40 metres. A path of width s metres is created around the playground for people to walk. Find an expression for the area of the path in terms of s.

Answer:
Side of playground = 40 m

A path of width s is made all around it.

So, side of larger square = 40 + 2s

Area of path = area of larger square - area of playground

= (40 + 2s)² - 40²
= 1600 + 160s + 4s² - 1600
= 4s² + 160s

Final answer:
Area of path = 4s² + 160s square metres

Question 8. If a number plus its reciprocal equals 10/3, find the number.

Answer:
Let the number be x.

Then:

x + 1/x = 10/3

Multiply by 3x:

3x² + 3 = 10x

3x² - 10x + 3 = 0

Factor:

3x² - 10x + 3
= 3x² - 9x - x + 3
= 3x(x - 3) - 1(x - 3)
= (3x - 1)(x - 3)

So:

(3x - 1)(x - 3) = 0

Therefore:

x = 1/3 or x = 3

Final answer:
The number is 3 or 1/3.

Question 9. A rectangular pool has area 2x² + 7x + 3 square hastas. If its width is 2x + 1 hastas, find its length.

Answer:
Area = length × width

Length = Area / Width

Length = (2x² + 7x + 3)/(2x + 1)

Factor numerator:

2x² + 7x + 3
= 2x² + 6x + x + 3
= 2x(x + 3) + 1(x + 3)
= (2x + 1)(x + 3)

So:

Length = [(2x + 1)(x + 3)]/(2x + 1)

Final answer:
Length = x + 3 hastas

Question 10. If both x - 2 and x - 1/2 are factors of px² + 5x + r, show that p = r.

Answer:
Let:

f(x) = px² + 5x + r

Since x - 2 is a factor:

f(2) = 0

p(2)² + 5(2) + r = 0
4p + 10 + r = 0 …(1)

Since x - 1/2 is a factor:

f(1/2) = 0

p(1/2)² + 5(1/2) + r = 0
p/4 + 5/2 + r = 0

Multiply by 4:

p + 10 + 4r = 0 …(2)

From (1):

4p + r = -10

From (2):

p + 4r = -10

Subtract:

4p + r = p + 4r
3p = 3r
p = r

Final answer:
p = r

Question 11. If a + b + c = 5 and ab + bc + ca = 10, prove that a³ + b³ + c³ - 3abc = -25.

Answer:
Use identity:

a³ + b³ + c³ - 3abc
= (a + b + c)(a² + b² + c² - ab - bc - ca)

Given:

a + b + c = 5
ab + bc + ca = 10

Now:

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

25 = a² + b² + c² + 20

So:

a² + b² + c² = 5

Now:

a² + b² + c² - ab - bc - ca
= 5 - 10
= -5

Therefore:

a³ + b³ + c³ - 3abc
= 5 × (-5)
= -25

Final answer:
a³ + b³ + c³ - 3abc = -25

Question 12. By factoring the expression, check that n³ - n is always divisible by 6 for all natural numbers n. Give reasons.

Answer:
n³ - n = n(n² - 1)
= n(n - 1)(n + 1)

So:

n³ - n = (n - 1)n(n + 1)

This is the product of three consecutive natural numbers.

Among any three consecutive natural numbers:

1. One number is divisible by 3.
2. At least one number is divisible by 2.

Therefore, their product is divisible by 2 × 3 = 6.

Final answer:
n³ - n is always divisible by 6.

Question 13. Find the value.

(i) x³ + y³ - 12xy + 64, when x + y = -4

Answer:
Use:

x³ + y³ = (x + y)³ - 3xy(x + y)

Given:

x + y = -4

x³ + y³ - 12xy + 64
= [(x + y)³ - 3xy(x + y)] - 12xy + 64
= (-4)³ - 3xy(-4) - 12xy + 64
= -64 + 12xy - 12xy + 64
= 0

Final answer:
0

(ii) x³ - 8y³ - 36xy - 216, when x = 2y + 6

Answer:
Given:

x = 2y + 6

So:

x - 2y = 6

Now:

x³ - 8y³ = x³ - (2y)³

Use:

a³ - b³ = (a - b)(a² + ab + b²)

x³ - (2y)³
= (x - 2y)(x² + 2xy + 4y²)

Since x - 2y = 6:

x³ - 8y³ - 36xy - 216
= 6(x² + 2xy + 4y²) - 36xy - 216
= 6x² + 12xy + 24y² - 36xy - 216
= 6x² - 24xy + 24y² - 216
= 6(x² - 4xy + 4y² - 36)
= 6[(x - 2y)² - 36]

Since x - 2y = 6:

= 6[6² - 36]
= 6(36 - 36)
= 0

Final answer:
0

Topics Covered in NCERT Solutions for Class 9 Maths Chapter 4

1. Algebraic identities
2. Difference between equations and identities
3. Visualising identities
4. Identity (a + b)²
5. Identity (a - b)²
6. Identity (a + b + c)²
7. Identity a² - b²
8. Factorisation using identities
9. Algebra tiles
10. Product of binomials
11. Factorisation of quadratic expressions
12. Splitting the middle term
13. Cube identities
14. Sum of cubes
15. Difference of cubes
16. Identity x³ + y³ + z³ - 3xyz
17. Rational algebraic expressions
18. Simplification by cancelling common factors
19. Area-based algebra questions
20. Proof-based identity questions

Important Formulas and Identities in NCERT Solutions for Class 9 Maths Chapter 4

Concept	Formula / Identity
Square of sum	(x + y)² = x² + 2xy + y²
Square of difference	(x - y)² = x² - 2xy + y²
Square of three terms	(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx
Difference of squares	(x + y)(x - y) = x² - y²
Product of binomials	(x + a)(x + b) = x² + (a + b)x + ab
General binomial product	(ax + b)(cx + d) = acx² + (ad + bc)x + bd
Difference of cubes	x³ - y³ = (x - y)(x² + xy + y²)
Sum of cubes	x³ + y³ = (x + y)(x² - xy + y²)
Cube of sum	(x + y)³ = x³ + 3x²y + 3xy² + y³
Cube of difference	(x - y)³ = x³ - 3x²y + 3xy² - y³
Three-variable cube identity	x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)

NCERT Class 9 Maths Ganita Manjari 2026 Chapter Solutions

Chapter	Title
Chapter 1	Orienting Yourself: The Use of Coordinates
Chapter 2	Introduction to Linear Polynomials
Chapter 3	The World of Numbers
Chapter 4	Exploring Algebraic Identities
Chapter 5	I’m Up and Down, and Round and Round
Chapter 6	Measuring Space: Perimeter and Area
Chapter 7	The Mathematics of Maybe: Introduction to Probability
Chapter 8	Predicting What Comes Next: Exploring Sequences and Progressions