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

RD Sharma Class 9 Solutions Chapter 5 Ex 5.4 Free Download

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, students will learn about factorisation of algebraic expressions of the form x3 + y3 + z3 – 3xyz.

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Access Answers to Maths RD Sharma 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= 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

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


Access other exercise solutions of Class 9 Maths Chapter 5 Factorization of Algebraic Expressions

Exercise 5.1 Solutions

Exercise 5.2 Solutions

Exercise 5.3 Solutions

Exercise VSAQs Solutions

RD Sharma Solutions for Class 9 Maths Chapter 5 Exercise 5.4

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 – 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

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