In this exercise of RD Sharma Chapter 5 Class 9 Maths Exercise 5.4 Solutions, students will learn about the Factorization of Algebraic Expressions. Students will get to know how to solve algebraic identities like- x3 + y3 + z3 – 3xyz. Also, about the Factorisation of the sum of the cubes when their sum is zero. Moreover, this exercise of Class 9 Chapter 5 of Algebraic Expressions will help to score easily in the exam as it is very easy to solve (get solve in 2 steps max), and formulae are not so difficult to learn.

Let’s go down and check the solution of the question, like- when the factorization of the sum of the cubes when their sum is zero. The attached PDF is prepared by our experts with an easy and explained step-by-step solution. Go through the article to know more about the Factorization of Algebraic Expressions.

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## Download RD Sharma Chapter 5 Class 9 Maths Exercise 5.4 Solutions

Class 9 Maths Chapter 5 Factorization of Algebraic Expressions Exercise 5.4

## Important Definitions RD Sharma Chapter 5 Class 9 Maths Exercise 5.4 Solutions

In this Exercise, the Factorization of Algebraic Expressions is based on the following topic-

- Factorization of the algebraic identity of the form x3 + y3 + z3 – 3xyz.

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

- Factorization of the sum of the cubes if their sum is zero.

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

On the basis of above agebraic expressions we have provided some examples below, although PDF is attached for more practice with lot of questions and their explanations-

**Ques- a****3**** + 8b****3 ****+ 64c****3 ****− 24abc**

**Solution**:

a3 + 8b3 + 64c3 − 24abc = (a)3 + (2b) 3 + (4c) 3− 3×a×2b×4c

Algebraic Identity- 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)

#### Ques- 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)

Algebraic Identity- 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)

#### Ques- (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)

## Frequently Asked Questions (FAQs) of RD Sharma Chapter 5 Class 9 Maths Exercise 5.4 Solutions

**Ques 1- On what topic of Exercise 5.4 of Chapter 5 Class 9 Algebraic Expression is based?**

**Ans- **Exercise 5.4 of Chapter 5 Class 9 is based on the Factorization of Algebraic Expression. Mainly the topics and related formulas are mentioned below-

- Factorization of the algebraic identity of the form x3 + y3 + z3 – 3xyz.

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

- Factorization of the sum of the cubes if their sum is zero.

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

**Ques 2- What is Algebraic Expressions?**

**Ans- **An algebraic expression is an identity formed up from integer constants, variables, and algebraic operations like- addition, multiplication, subtraction, division, and exponentiation by a type that is a rational number).

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