Chapter 4: Factorisation — Exercise 4(C) | Solutions for Class 9 RS Aggarwal ICSE Foundation Mathematics

Exercise 4C

Question 1

Factorise:

x³ + 64

Answer

Given,

⇒ x³ + 64

⇒ (x)³ + (4)³

By using the identity,

a³ + b³ = (a + b)(a² - ab + b²)

⇒ (x + 4)(x² - x × 4 + 4²)

⇒ (x + 4)(x² - 4x + 16)

Hence, x³ + 64 = (x + 4)(x² - 4x + 16).

Question 2

Factorise:

8a³ + 27b³

Answer

Given,

⇒ 8a³ + 27b³

⇒ (2a)³ + (3b)³

By using the identity,

a³ + b³ = (a + b)(a² - ab + b²)

⇒ (2a + 3b)[(2a)² - 2a × 3b + (3b)²]

⇒ (2a + 3b)(4a² - 6ab + 9b²)

Hence, 8a³ + 27b³ = (2a + 3b)(4a² - 6ab + 9b²).

Question 3

Factorise:

7a³ + 56b³

Answer

Given,

⇒ 7a³ + 56b³

⇒ 7(a³ + 8b³)

⇒ 7[a³ + (2b)³]

By using the identity,

a³ + b³ = (a + b)(a² - ab + b²)

⇒ 7(a + 2b)(a² - a × 2b + (2b)²)

⇒ 7(a + 2b)(a² - 2ab + 4b²).

Hence, 7a³ + 56b³ = 7(a + 2b)(a² - 2ab + 4b²).

Question 4

Factorise:

x⁵ + x²

Answer

Given,

⇒ x⁵ + x²

⇒ x²(x³ + 1)

⇒ x²[(x)³ + (1)³]

By using the identity,

a³ + b³ = (a + b)(a² - ab + b²)

⇒ x²(x + 1)(x² - x × 1 + 1²)

⇒ x²(x + 1)(x² - x + 1).

Hence, x⁵ + x² = x²(x + 1)(x² - x + 1).

Question 5

Factorise:

16x⁴ + 54x

Answer

Given,

⇒ 16x⁴ + 54x

⇒ 2x(8x³ + 27)

⇒ 2x[(2x)³ + (3)³]

By using the identity,

a³ + b³ = (a + b)(a² - ab + b²)

⇒ 2x(2x + 3)[(2x)² - 2x × 3 + (3)²]

⇒ 2x(2x + 3)(4x² - 6x + 9).

Hence, 16x⁴ + 54x = 2x(2x + 3)(4x² - 6x + 9).

Question 6

Factorise:

$216x^3 + \dfrac{1}{27}$

Answer

Given,

⇒ $216x^3 + \dfrac{1}{27}$

By using the identity,

a³ + b³ = (a + b)(a² - ab + b²)

$\Rightarrow (6x)^3 + \Big(\dfrac{1}{3}\Big)^3 \\[1em] \Rightarrow \Big(6x + \dfrac{1}{3}\Big)\Big[(6x)^2 - 6x \times \Big(\dfrac{1}{3}\Big) + \Big(\dfrac{1}{3}\Big)^2\Big] \\[1em] \Rightarrow \Big(6x + \dfrac{1}{3}\Big)\Big(36x^2 - 2x + \dfrac{1}{9}\Big).$

Hence, $216x^3 + \dfrac{1}{27} = \Big(6x + \dfrac{1}{3}\Big)\Big(36x^2 - 2x + \dfrac{1}{9}\Big)$ .

Question 7

Factorise:

a⁶ + b⁶

Answer

Given,

⇒ a⁶ + b⁶

⇒ (a²)³ + (b²)³

⇒ (a² + b²)[(a²)² - a² × b² + (b²)²]

⇒ (a² + b²)(a⁴ - a²b² + b⁴)

Hence, a⁶ + b⁶ = (a² + b²)(a⁴ - a²b² + b⁴).

Question 8

Factorise:

a⁴ + 343a

Answer

Given,

⇒ a⁴ + 343a

⇒ a(a³ + 343)

⇒ a(a³ + 7³)

By using the identity,

a³ + b³ = (a + b)(a² - ab + b²)

⇒ a(a + 7)[(a)² - a × 7 + (7)²]

⇒ a(a + 7)(a² - 7a + 49)

Hence, a⁴ + 343a = a(a + 7)(a² - 7a + 49).

Question 9

Factorise:

125x³ + 1

Answer

Given,

⇒ 125x³ + 1

⇒ (5x)³ + (1)³

By using the identity,

a³ + b³ = (a + b)(a² - ab + b²)

⇒ (5x + 1)[(5x)² - 5x × 1 + (1)²]

⇒ (5x + 1)(25x² - 5x + 1)

Hence, 125x³ + 1 = (5x + 1)(25x² - 5x + 1).

Question 10

Factorise:

2a³ + 16b³ - 3a - 6b

Answer

Given,

⇒ 2a³ + 16b³ - 3a - 6b

⇒ (2a³ + 16b³) - (3a + 6b)

⇒ 2(a³ + 8b³) - 3(a + 2b)

⇒ 2[(a)³ + (2b)³] - 3(a + 2b)

⇒ 2(a + 2b)[(a)² - a × 2b + (2b)²] - 3(a + 2b)

⇒ 2(a + 2b)(a² - 2ab + 4b²) - 3(a + 2b)

⇒ (a + 2b)[2(a² - 2ab + 4b²) - 3]

Hence, 2a³ + 16b³ - 3a - 6b = (a + 2b)[2(a² - 2ab + 4b²) - 3].

Question 11

Factorise:

a³ - 125 - 2a + 10

Answer

Given,

⇒ a³ - 125 - 2a + 10

⇒ (a³ - 125) - 2a + 10

⇒ [(a)³ - (5)³] - 2(a - 5)

By using the identity,

a³ - b³ = (a - b)(a² + ab + b²)

⇒ (a - 5)(a² + a × 5 + 5²) - 2(a - 5)

⇒ [(a - 5)(a² + 5a + 25)] - 2(a - 5)

⇒ (a - 5)[(a² + 5a + 25) - 2]

⇒ (a - 5)(a² + 5a + 23)

Hence, a³ - 125 - 2a + 10 = (a - 5)(a² + 5a + 23).

Question 12

Factorise:

x³ - 125

Answer

Given,

⇒ x³ - 125

⇒ (x)³ - (5)³

By using the identity,

a³ - b³ = (a - b)(a² + ab + b²)

⇒ (x - 5)(x² + x × 5 + 5²)

⇒ (x - 5)(x² + 5x + 25).

Hence, x³ - 125 = (x - 5)(x² + 5x + 25).

Question 13

Factorise:

$8a^3 - \dfrac{1}{27b^3}$

Answer

Given,

$\Rightarrow 8a^3 - \dfrac{1}{27b^3} \\[1em] \Rightarrow (2a)^3 - \Big(\dfrac{1}{3b}\Big)^3 \\[1em] \Rightarrow \Big(2a - \dfrac{1}{3b}\Big)\Big[(2a)^2 + 2a \times \Big(\dfrac{1}{3b}\Big) + \Big(\dfrac{1}{3b}\Big)^2\Big] \\[1em] \Rightarrow \Big(2a - \dfrac{1}{3b}\Big)\Big(4a^2 + \dfrac{2a}{3b} + \dfrac{1}{9b^2}\Big).$

Hence, $8a^3 - \dfrac{1}{27b^3} = \Big(2a - \dfrac{1}{3b}\Big)\Big(4a^2 + \dfrac{2a}{3b} + \dfrac{1}{9b^2}\Big)$ .

Question 14

Factorise:

$\dfrac{8a^3}{27} - \dfrac{b^3}{8}$

Answer

$\Rightarrow \dfrac{8a^3}{27} - \dfrac{b^3}{8} \\[1em] \Rightarrow \Big(\dfrac{2a}{3}\Big)^3 - \Big(\dfrac{b}{2}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{2a}{3} - \dfrac{b}{2}\Big)\Big[\Big(\dfrac{2a}{3}\Big)^2 + \Big(\dfrac{2a}{3}\Big) \times \Big(\dfrac{b}{2}\Big) + \Big(\dfrac{b}{2}\Big)^2\Big] \\[1em] \Rightarrow \Big(\dfrac{2a}{3} - \dfrac{b}{2}\Big)\Big(\dfrac{4a^2}{9} + \dfrac{2ab}{6} + \dfrac{b^2}{4}\Big) \\[1em] \Rightarrow \Big(\dfrac{2a}{3} - \dfrac{b}{2}\Big)\Big(\dfrac{4a^2}{9} + \dfrac{ab}{3} + \dfrac{b^2}{4}\Big).$

Hence, $\dfrac{8a^3}{27} - \dfrac{b^3}{8} = \Big(\dfrac{2a}{3} - \dfrac{b}{2}\Big)\Big(\dfrac{4a^2}{9} + \dfrac{ab}{3} + \dfrac{b^2}{4}\Big)$ .

Question 15

Factorise:

a - 8ab³

Answer

Given,

⇒ a - 8ab³

⇒ a(1 - 8b³)

⇒ a[(1)³ - (2b)³]

By using the identity,

(a³ - b³) = (a - b)(a² + ab + b²)

⇒ a(1 - 2b)[(1)² + 1 × 2b + (2b)²]

⇒ a(1 - 2b)(1 + 2b + 4b²).

Hence, a - 8ab³ = a(1 - 2b)(1 + 2b + 4b²).

Question 16

Factorise:

x⁶ - 1

Answer

Given,

⇒ x⁶ - 1

⇒ (x³)² - (1)²

By using the identity,

(a² - b²) = (a + b)(a - b)

⇒ (x³ - 1)(x³ + 1)

⇒ [x³ - (1)³][x³ + (1)³]

⇒ (x - 1)[(x)² + x × 1 + 1²] [(x + 1)(x)² - x × 1 + 1²]

⇒ (x - 1)(x² + x + 1)(x + 1)(x² - x + 1)

⇒ (x - 1)(x + 1)(x² + x + 1)(x² - x + 1).

Hence, x⁶ - 1 = (x - 1)(x + 1)(x² + x + 1)(x² - x + 1).

Question 17

Factorise:

a³ - 0.064

Answer

Given,

⇒ a³ - 0.064

⇒ (a)³ - (0.4)³

By using the identity,

(a³ - b³) = (a - b)(a² + ab + b²)

⇒ (a - 0.4)[a² + 0.4a + (0.4)²]

⇒ (a - 0.4)(a² + 0.4a + 0.16).

Hence, a³ - 0.064 = (a - 0.4)(a² + 0.4a + 0.16).

Question 18

Factorise:

24x⁴ - 375x

Answer

Given,

⇒ 24x⁴ - 375x

⇒ 3x(8x³ - 125)

⇒ 3x[(2x)³ - (5)³]

By using the identity,

(a³ - b³) = (a - b)(a² + ab + b²)

⇒ 3x(2x - 5)[(2x)² + 2x × 5 + (5)²]

⇒ 3x(2x - 5)(4x² + 10x + 25).

Hence, 24x⁴ - 375x = 3x(2x - 5)(4x² + 10x + 25).

Question 19

Factorise:

3a⁷b - 81a⁴b⁴

Answer

Given,

⇒ 3a⁷b - 81a⁴b⁴

⇒ 3a⁴b(a³ - 27b³)

⇒ 3a⁴b[a³ - (3b)³]

⇒ 3a⁴b(a - 3b)[(a)² + a × 3b + (3b)²]

⇒ 3a⁴b(a - 3b)(a² + 3ab + 9b²).

Hence, 3a⁷b - 81a⁴b⁴ = 3a⁴b(a - 3b)(a² + 3ab + 9b²).

Question 20

Factorise:

$a^3 - \dfrac{1}{a^3} - 2a + \dfrac{2}{a}$

Answer

Given,

$\Rightarrow a^3 - \dfrac{1}{a^3} - 2a + \dfrac{2}{a} \\[1em] \Rightarrow a^3 - \dfrac{1}{a^3} - 2\Big( a - \dfrac{1}{a}\Big) \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)\Big[(a)^2 + 1 \times a \times \dfrac{1}{a} + \Big(\dfrac{1}{a}\Big)^2\Big] - 2\Big( a - \dfrac{1}{a}\Big) \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big) \Big(a^2 + 1 + \dfrac{1}{a^2}\Big)- 2\Big( a - \dfrac{1}{a}\Big) \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big) \Big[\Big(a^2 + 1 + \dfrac{1}{a^2}\Big)- 2\Big] \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big) \Big(a^2 + \dfrac{1}{a^2} - 1\Big).$

Hence, $a^3 - \dfrac{1}{a^3} - 2a + \dfrac{2}{a} = \Big(a - \dfrac{1}{a}\Big) \Big(a^2 + \dfrac{1}{a^2} - 1\Big)$ .

Question 21

Factorise:

2x⁷ - 128x

Answer

Given,

⇒ 2x⁷ - 128x

⇒ 2x(x⁶ - 64)

⇒ 2x[(x³)² - (8)²]

⇒ 2x[(x)³ - (8)] [(x)³ + (8)]

⇒ 2x[x³ - 8] [x³ + 8]

⇒ 2x[x³ - 2³] [x³ + 2³]

By using the identity,

(a³ - b³) = (a - b)(a² + ab + b²)

(a³ + b³) = (a + b)(a² - ab + b²)

⇒ 2x[(x - 2)(x² + 2 × x + (2)²)] [(x + 2)(x)² - 2 × x + (2)²]

⇒ 2x(x - 2)(x² + 2x + 4)(x + 2)(x² - 2x + 4)

⇒ 2x(x - 2)(x + 2)(x² + 2x + 4)(x² - 2x + 4).

Hence, 2x⁷ - 128x = 2x(x - 2)(x + 2)(x² + 2x + 4)(x² - 2x + 4).

Question 22

Factorise:

250(a - b)³ + 2

Answer

Given,

⇒ 250(a - b)³ + 2

⇒ 2[125(a - b)³ + 1]

⇒ 2{[5(a - b)]³ + (1)³}

By using the identity,

(a³ + b³) = (a + b)(a² - ab + b²)

⇒ 2[5(a - b) + 1] {[5(a - b)]² - 5(a - b) × 1 + (1)²}

⇒ 2[(5a - 5b + 1)(25(a - b)² - 5(a - b) + 1)]

⇒ 2[(5a - 5b + 1)(25(a² - 2ab + b²) - 5a + 5b + 1)]

⇒ 2[(5a - 5b + 1)(25a² - 50ab + 25b² - 5a + 5b + 1)]

Hence, 250(a - b)³ + 2 = 2[(5a - 5b + 1)(25a² - 50ab + 25b² - 5a + 5b + 1)].

Question 23

Factorise:

8a³ - b³ - 4ax + 2bx

Answer

Given,

⇒ 8a³ - b³ - 4ax + 2bx

⇒ (2a)³ - (b)³ - 2x(2a - b)

By using the identity,

(a³ - b³) = (a - b)(a² + ab + b²)

⇒ (2a - b)[(2a)² + 2a × b + (b)²] - 2x(2a - b)

⇒ (2a - b)(4a² + 2ab + b²) - 2x(2a - b)

⇒ (2a - b)(4a² + 2ab + b² - 2x).

Hence, 8a³ - b³ - 4ax + 2bx = (2a - b)(4a² + 2ab + b² - 2x).

Question 24

Factorise:

a³ - 27b³ + 2a²b - 6ab²

Answer

Given,

⇒ a³ - 27b³ + 2a²b - 6ab²

⇒ (a)³ - (3b)³ + 2ab(a - 3b)

By using the identity,

(a³ - b³) = (a - b)(a² + ab + b²)

⇒ (a - 3b)[a² + a × 3b + (3b)²] + 2ab(a - 3b)

⇒ (a - 3b)[a² + 3ab + 9b²] + 2ab(a - 3b)

⇒ (a - 3b)(a² + 3ab + 9b² + 2ab)

⇒ (a - 3b)(a² + 5ab + 9b²).

Hence, a³ - 27b³ + 2a²b - 6ab² = (a - 3b)(a² + 5ab + 9b²).

Question 25

Factorise:

32a²x³ - 8b²x³ - 4a²y³ + b²y³

Answer

Given,

⇒ 32a²x³ - 8b²x³ - 4a²y³ + b²y³

⇒ 8x³(4a² - b²) - y³(4a² - b²)

⇒ (8x³ - y³)(4a² - b²)

⇒ (4a² - b²)[(2x)³ - (y)³]

By using the identity,

(a³ - b³) = (a - b)(a² + ab + b²)

⇒ (4a² - b²){(2x - y)[(2x)² + 2x × y + (y)²]}

⇒ [(2a)² - (b)²][(2x - y)(4x² + 2xy + y²)]

By using the identity,

(a² - b²) = (a + b)(a - b)

⇒ [(2a)² - (b)²] [(2x - y)(4x² + 2xy + y²)]

⇒ (2a + b)(2a - b)(2x - y)(4x² + 2xy + y²).

Hence, 32a²x³ - 8b²x³ - 4a²y³ + b²y³ = (2a + b)(2a - b)(2x - y)(4x² + 2xy + y²).

Question 26

Factorise:

a² - 4b² + a³ - 8b³ - (a - 2b)²

Answer

Given,

⇒ a² - 4b² + a³ - 8b³ - (a - 2b)²

⇒ (a)² - (2b)² + (a)³ - (2b)³ - (a - 2b)(a - 2b)

By using the identity,

(a² - b²) = (a + b)(a - b) and (a³ - b³) = (a - b)(a² + ab + b²)

⇒ (a + 2b)(a - 2b) + (a - 2b)(a² + a × 2b + (2b)²) - (a - 2b)(a - 2b)

⇒ (a - 2b)[(a + 2b) + (a² + 2ab + 4b²) - (a - 2b)]

⇒ (a - 2b)[a + 2b - a + 2b + (a² + 2ab + 4b²)]

⇒ (a - 2b)(a² + 2ab + 4b² + 4b).

Hence, a² - 4b² + a³ - 8b³ - (a - 2b)² = (a - 2b)(a² + 2ab + 4b² + 4b).

Question 27

Factorise:

(a + b)³ + (a - b)³

Answer

Given,

⇒ (a + b)³ + (a - b)³

By using the identity,

(a³ + b³) = (a + b)(a² - ab + b²)

⇒ [(a + b) + (a - b)] [(a + b)² - (a + b) × (a - b) + (a - b)²]

⇒ (2a)[a² + 2ab + b² - (a² - b²) + a² - 2ab + b²]

⇒ (2a)[a² + 2ab + b² - a² + b² + a² - 2ab + b²]

⇒ (2a)(a² + 3b²).

Hence, (a + b)³ + (a - b)³ =(2a)(a² + 3b²).

Question 28

Factorise:

x³ - 3x² + 3x + 7

Answer

Given,

⇒ x³ - 3x² + 3x + 7

⇒ x³ - 3x² + 3x + 8 - 1

⇒ x³ + 8 - 3x² + 3x - 1

⇒ x³ - 1³ - 3 × 1 x (x + 1) + 8

⇒ (x - 1)³ + 2³

By using the identity,

(a³ + b³) = (a + b)(a² - ab + b²)

⇒ [(x - 1) + 2] [(x - 1)² - (x - 1) × 2 + 2²]

⇒ (x + 1)[(x² - 2x + 1) - (2x - 2) + 4]

⇒ (x + 1)[(x² - 2x + 1) - 2x + 2 + 4]

⇒ (x + 1)(x² - 4x + 7)

Hence, x³ - 3x² + 3x + 7 = (x + 1)(x² - 4x + 7).

Chapter 4

Factorisation — Exercise 4(C)

Class - 9 RS Aggarwal Mathematics Solutions

Exercise 4C

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22

Question 23

Question 24

Question 25

Question 26

Question 27

Question 28