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

RD Sharma Class 9 Mathematics Chapter 5 Exercise 5.4 Factorization of Algebraic Expressions is 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 factorisation of algebraic expressions of the form x3 + y3 + z3 – 3xyz.

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#### Exercise 5.4 Page No: 5.22

Factorize each of the following expressions:

Question 1: a3 + 8b3 + 64c3 âˆ’ 24abc

Solution:

a3 + 8b3 + 64c3 âˆ’ 24abc

= (a)3 + (2b) 3 + (4c) 3âˆ’ 3Ã—aÃ—2bÃ—4c

[Using a3+b3+c3âˆ’3abc = (a+b+c)(a2+b2+c2âˆ’abâˆ’bcâˆ’ca)]

= (a+2b+4c)(a2+(2b)2 + (4c)2âˆ’aÃ—2bâˆ’2bÃ—4câˆ’4cÃ—a)

= (a+2b+4c)(a2 +4b2 +16c2 âˆ’2abâˆ’8bcâˆ’4ac)

Therefore, a3 + 8b3 + 64c3 âˆ’ 24abc = (a+2b+4c)(a2 +4b2 +16c2 âˆ’2abâˆ’8bcâˆ’4ac)

Question 2: x 3 âˆ’ 8y 3+ 27z3 + 18xyz

Solution:

= x3 âˆ’ (2y) 3 + (3z) 3 âˆ’ 3Ã—xÃ—(âˆ’2y)(3z)

= (x + (âˆ’2y) + 3z) (x2 + (âˆ’2y)2 + (3z) 2 âˆ’x(âˆ’2y)âˆ’(âˆ’2y)(3z)âˆ’3z(x))

[using a3+b3+c3âˆ’3abc = (a+b+c)(a2+b2+c2âˆ’abâˆ’bcâˆ’ca)]

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

Question 3: 27x 3 âˆ’ y 3– z3 – 9xyz

Solution:

27x 3 âˆ’ y 3– z3 – 9xyz

= (3x) 3 âˆ’ y 3– z3 â€“ 3(3xyz)

[Using a3 + b3 + c3 âˆ’3abc = (a + b + c)(a2+b2+c2âˆ’abâˆ’bcâˆ’ca)]

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

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

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

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

Solution:

1/27 x3 âˆ’ y3 + 125z3 + 5xyz

= (x/3)3+(âˆ’y)3 +(5z)3 â€“ 3 x/3 (âˆ’y)(5z)

[Using a3 + b3 + c3 âˆ’3abc = (a + b + c)(a2+b2+c2âˆ’abâˆ’bcâˆ’ca)]

= (x/3 + (âˆ’y) + 5z)((x/3)2 + (âˆ’y)2 + (5z) 2 â€“x/3(âˆ’y) âˆ’ (âˆ’y)5zâˆ’5z(x/3))

= (x/3 âˆ’y + 5z) (x^2/9 + y2 + 25z2 + xy/3 + 5yz â€“ 5zx/3)

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

Solution:

8x3 + 27y3 âˆ’ 216z3 + 108xyz

= (2x) 3 + (3y) 3 +(âˆ’6y) 3 âˆ’3(2x)(3y)(âˆ’6z)

= (2x+3y+(âˆ’6z)){ (2x)2+(3y) 2+(âˆ’6z) 2 âˆ’2xÃ—3yâˆ’3y(âˆ’6z)âˆ’(âˆ’6z)2x}

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

Question 6: 125 + 8x3 âˆ’ 27y3 + 90xy

Solution:

125 + 8x3 âˆ’ 27y3 + 90xy

= (5)3 + (2x) 3 +(âˆ’3y) 3 âˆ’3Ã—5Ã—2xÃ—(âˆ’3y)

= (5+2x+(âˆ’3y)) (52 +(2x) 2 +(âˆ’3y) 2 âˆ’5(2x)âˆ’2x(âˆ’3y)âˆ’(âˆ’3y)5)

= (5+2xâˆ’3y)(25+4x2 +9y2 âˆ’10x+6xy+15y)

Question 7: (3xâˆ’2y)3 + (2yâˆ’4z) 3 + (4zâˆ’3x) 3

Solution:

(3xâˆ’2y)3 + (2yâˆ’4z) 3 + (4zâˆ’3x) 3

Let (3xâˆ’2y) = a, (2yâˆ’4z) = b , (4zâˆ’3x) = c

a + b + c= 3xâˆ’2y+2yâˆ’4z+4zâˆ’3x = 0

We know, a3 + b3 + c3 âˆ’3abc = (a + b + c)(a2+b2+c2âˆ’abâˆ’bcâˆ’ca)

â‡’ a3 + b3 + c3 âˆ’3abc = 0

or a3 + b3 + c3 =3abc

â‡’ (3xâˆ’2y)3 + (2yâˆ’4z) 3 + (4zâˆ’3x) 3 = 3(3xâˆ’2y)(2yâˆ’4z)(4zâˆ’3x)

Question 8: (2xâˆ’3y)3 + (4zâˆ’2x) 3 + (3yâˆ’4z) 3

Solution:

(2xâˆ’3y)3 + (4zâˆ’2x) 3 + (3yâˆ’4z) 3

Let 2x â€“ 3y = a , 4z â€“ 2x = b , 3y â€“ 4z = c

a + b + c= 2xâˆ’3y+4zâˆ’2x+3yâˆ’4z = 0

We know, a3 + b3 + c3 âˆ’3abc = (a + b + c)(a2+b2+c2âˆ’abâˆ’bcâˆ’ca)

â‡’ a3 + b3 + c3 âˆ’3abc = 0

(2xâˆ’3y)3 + (4zâˆ’2x) 3 + (3yâˆ’4z) 3 = 3(2xâˆ’3y)(4zâˆ’2x)(3yâˆ’4z)

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

RD Sharma SolutionsÂ Class 9 Maths Chapter 5 Factorization of Algebraic Expressions Exercise 5.4 is based on the following topics:

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

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

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

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