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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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
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