# RD Sharma Solutions for Class 9 Maths Chapter 5 Factorization of Algebraic Expressions Exercise 5.4

RD Sharma Solutions for Class 9 Mathematics Chapter 5 Exercise 5.4 Factorization of Algebraic Expressions are provided here. The Class 9 Maths Chapter 5 deals with the topic of algebraic expressions. Algebraic Expression in Mathematics is an expression built from algebraic operations. In this exercise of RD Sharma Solutions Class 9, students will learn about the factorization of algebraic expressions of form x3 + y3 + z3 – 3xyz.

## RD Sharma Solutions for Class 9 Maths Chapter 5 Factorization of Algebraic Expressions Exercise 5.4

### Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 5 Factorization of Algebraic Expressions Exercise 5.4 Page Number 5.22

#### Exercise 5.4 Page No: 5.22

Factorize each of the following expressions:

Question 1: a3 + 8b3 + 64c3 − 24abc

Solution:

a3 + 8b3 + 64c3 − 24abc

= (a)3 + (2b) 3 + (4c) 3− 3×a×2b×4c

[Using a3+b3+c3−3abc = (a+b+c)(a2+b2+c2−ab−bc−ca)]

= (a+2b+4c)(a2+(2b)2 + (4c)2−a×2b−2b×4c−4c×a)

= (a+2b+4c)(a2 +4b2 +16c2 −2ab−8bc−4ac)

Therefore, a3 + 8b3 + 64c3 − 24abc = (a+2b+4c)(a2 +4b2 +16c2 −2ab−8bc−4ac)

Question 2: x 3 − 8y 3+ 27z3 + 18xyz

Solution:

= x3 − (2y) 3 + (3z) 3 − 3×x×(−2y)(3z)

= (x + (−2y) + 3z) (x2 + (−2y)2 + (3z) 2 −x(−2y)−(−2y)(3z)−3z(x))

[using a3+b3+c3−3abc = (a+b+c)(a2+b2+c2−ab−bc−ca)]

=(x −2y + 3z)(x2 + 4y2 + 9 z2 + 2xy + 6yz − 3zx)

Question 3: 27x 3 − y 3– z3 – 9xyz

Solution:

27x 3 − y 3– z3 – 9xyz

= (3x) 3 − y 3– z3 – 3(3xyz)

[Using a3 + b3 + c3 −3abc = (a + b + c)(a2+b2+c2−ab−bc−ca)]

Here a = 3x, b = -y and c = -z

= (3x – y – z){ (3x)2 + (- y)2 + (– z)2 + 3xy – yz + 3xz)}

= (3x – y – z){ 9x2 + y2 + z2 + 3xy – yz + 3xz)}

Question 4: 1/27 x3 − y3 + 125z3 + 5xyz

Solution:

1/27 x3 − y3 + 125z3 + 5xyz

= (x/3)3+(−y)3 +(5z)3 – 3 x/3 (−y)(5z)

[Using a3 + b3 + c3 −3abc = (a + b + c)(a2+b2+c2−ab−bc−ca)]

= (x/3 + (−y) + 5z)((x/3)2 + (−y)2 + (5z) 2 –x/3(−y) − (−y)5z−5z(x/3))

= (x/3 −y + 5z) (x^2/9 + y2 + 25z2 + xy/3 + 5yz – 5zx/3)

Question 5: 8x3 + 27y3 − 216z3 + 108xyz

Solution:

8x3 + 27y3 − 216z3 + 108xyz

= (2x) 3 + (3y) 3 +(−6y) 3 −3(2x)(3y)(−6z)

= (2x+3y+(−6z)){ (2x)2+(3y) 2+(−6z) 2 −2x×3y−3y(−6z)−(−6z)2x}

= (2x+3y−6z) {4x2 +9y2 +36z2 −6xy + 18yz + 12zx}

Question 6: 125 + 8x3 − 27y3 + 90xy

Solution:

125 + 8x3 − 27y3 + 90xy

= (5)3 + (2x) 3 +(−3y) 3 −3×5×2x×(−3y)

= (5+2x+(−3y)) (52 +(2x) 2 +(−3y) 2 −5(2x)−2x(−3y)−(−3y)5)

= (5+2x−3y)(25+4x2 +9y2 −10x+6xy+15y)

Question 7: (3x−2y)3 + (2y−4z) 3 + (4z−3x) 3

Solution:

(3x−2y)3 + (2y−4z) 3 + (4z−3x) 3

Let (3x−2y) = a, (2y−4z) = b , (4z−3x) = c

a + b + c= 3x−2y+2y−4z+4z−3x = 0

We know, a3 + b3 + c3 −3abc = (a + b + c)(a2+b2+c2−ab−bc−ca)

⇒ a3 + b3 + c3 −3abc = 0

or a3 + b3 + c3 =3abc

⇒ (3x−2y)3 + (2y−4z) 3 + (4z−3x) 3 = 3(3x−2y)(2y−4z)(4z−3x)

Question 8: (2x−3y)3 + (4z−2x) 3 + (3y−4z) 3

Solution:

(2x−3y)3 + (4z−2x) 3 + (3y−4z) 3

Let 2x – 3y = a , 4z – 2x = b , 3y – 4z = c

a + b + c= 2x−3y+4z−2x+3y−4z = 0

We know, a3 + b3 + c3 −3abc = (a + b + c)(a2+b2+c2−ab−bc−ca)

⇒ a3 + b3 + c3 −3abc = 0

(2x−3y)3 + (4z−2x) 3 + (3y−4z) 3 = 3(2x−3y)(4z−2x)(3y−4z)

## RD Sharma Solutions for Class 9 Maths Chapter 5 Factorization of Algebraic Expressions Exercise 5.4

RD Sharma Solutions for Class 9 Maths Chapter 5 Factorization of Algebraic Expressions Exercise 5.4 are based on the following topics:

1. Factorization of algebraic expressions of form x3 + y3 + z3 – 3xyz

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

2. Factorization of the sum of the cubes when their sum is zero

If (x + y + z) = 0 then x3 + y3 + z3 = 3xyz