RD Sharma Chapter 5 Class 9 Maths Exercise 5.2 Solutions

In this Chapter of Factorization of Algebraic Expression, students will learn about the Factorization of algebraic identities expressible as the sum or difference of two cubes. These identities will be explained step-by-step in the following PDF attached to this article. RD Sharma Chapter 5 Class 9 Maths Exercise 5.2 Solutions are prepared by the professional by keeping in mind the shortcuts and easy methods, which helps learners to understand the simplification easily.

The problems mentioned in the RD Sharma Chapter 5 Class 9 Maths Exercise 5.2 Solutions PDF are the combination of the questions given in the RD Sharma and CBSE. Students should by-heart the formulas based on the Factorization of Algebraic Expressions the sum and differences of two cubes while going through the solving problems.

Learn about RD Sharma Chapter 5 (Factorization of Algebraic Expressions) Class 9

Download RD Sharma Chapter 5 Class 9 Maths Exercise 5.2 Solutions PDF

Class 9 Maths Chapter 5 Factorization of Algebraic Expressions Exercise 5.2

Important Definitions RD Sharma Chapter 5 Class 9 Maths Exercise 5.2 Solutions

To memorize the patterns involved in the formula of Sum and Difference of two cubes is given in the below points-

1. The polynomial in the form x3+y3 is known as the sum of two cubes because two cubic terms getting added together.
2. The polynomial in the form x3-y3 is known as the difference of two cubes because two cubic terms are signifying deduction.

So, here are the formulas that summarize about the sum and difference of two cubes-

• a3 + b3 = (a + b) (a2 + b2 – ab)
• a3 – b3 = (a – b) (a2 + b2 + ab)

Follow the steps while doing factorization of algebraic identities-

1. “Factor Out” any of the common terms.
2. Check if it fits any of the identities, plus any more may know.
3. Keep going till can’t factor the identities anymore.

We have provided some examples below in which the factorization of algebraic expressions is solved in expanded form-

Ques- 8x3y3 + 27a3

Solution- 8x3y3+27a3 =(2xy)3+(3a)3

Algebraic Identity- a3 + b3 = (a + b) (a2 + b2 – ab)

= 8x3y3+27a3 =(2xy+3a) {(2xy)2-2xy3a+(3a)2}

= (2xy+3a) (4x2y2-6ax+9a2)

Therefore, the required factorization of 8x3y3+27a3 is (2xy+3a) (4x2y2-6ax+9a2)

Ques- 64a3 − b3

Solution- 64a3 − b3 = (4a)3− b3

Algebraic Identity- a3 – b3 = (a – b) (a2 + ab + b2)

64a3 − b3 = (4a−b) {(4a)2 + 4a×b + b2}

=(4a−b) (16a2 +4ab+b2)

Therefore, the required factorization of 64a3 − b3 is (4a−b) (16a2 +4ab+b2)

Frequently Asked Questions (FAQs) of RD Sharma Chapter 5 Class 9 Maths Exercise 5.2 Solutions

Ques 1- How do you factor the sum or difference of two cubes?

Ans- The sum or difference of two cubes can be factorized into an output of binomial times a trinomial.

Ques 2- Which expression is the difference between two cubes?

Ans- The difference of two cubes is equivalent to the difference of their cube roots times a trinomial, which comprises the squares of the cube roots and the reverse of the outcome of the cube roots. A number’s opposite is that likewise number among a distinctive sign in front.