# RD Sharma Chapter 4 Class 9 Maths Exercise 4.3 Solutions

RD Sharma Chapter 4 Class 9 Maths Exercise 4.3 Solutions is about the cube of a binomial. In this article, the identity will get extended for the square of a binomial to the cube of a binomial. The PDF is attached to this article for the learners, which helps them solve the problems with an easy method and shortcuts provided by our experts.

In this article, we have attached the RD Sharma Chapter 4 Class 9 Maths Exercise 4.3 Solutions PDF for the students, which helps them to practice to score well in the exam. The students will learn how to solve the questions in a limited time with shortcuts. Basically, in this exercise, it is very important to remember the formulas.

Learn about RD Sharma Class 9 Chapter 4- Algebraic Identities

## Download RD Sharma Chapter 4 Class 9 Maths Exercise 4.3 Solutions PDF

Solutions for Class 9 Maths Chapter 4 Algebraic Identities Exercise 4.3

## Important Definitions RD Sharma Chapter 4 Class 9 Maths Exercise 4.3 Solutions

The questions of this exercise will be solved based on the formulas mentioned in the following points-

### Sum of Cubes

(x + y)3= x3 + 3x2y + 3xy2 + y3

= x3 + 3xy(x + y) + y3

### Differences of Cubes

(x – y)3 = x3 – 3x2y + 3xy2 – y3

= x3 – 3xy(x – y) – y3

Here are some examples we are showing for the learners to understand about applying the formula in a correct question-

Ques)- (p + 5q)3 + (p – 5q)3

Solution-

Algebraic Identity- (a+b)3 = a3 + b3 + 3ab(a+b)

= p3 + 3(p2 x 5q) + 3(p x 5q2)+ (5q)3 + p3 – 3(p2 x 5q) + 3(p x 5q2) – (5q)3

= x3 + 15p2q + 75pq2 + 125q3 + p3 – 15p2q + 75pq2 – 125q3

= 2p3 + 150pq2

Therefore, (p + 5q)3 + (p – 5q)3 = 2p3 + 150pq2

Ques)- (5m + 2n)3 – (5m – 2n)3

Solution-

= (5m + 2n)3 – (5m – 2n)3

= {(5m)3 + 3 (5m)2 (2n) + 3 (5m) (2n)2 + (2n)3} – {(5m)3 – 3 (5m)2 (2n) + 3 (5m) (2n)2 – (2n)3}

= {125 m3 + 150 m2 n + 60 m n2 + 8 n3} – {125 m3 – 150 m2 n + 60 m n2 – 8 n3}

= 125 m3 + 150 m2 n + 60 m n2 + 8 n3 – 125 m3 + 150 m2 n – 60 m n2 + 8 n3

= 125 m3 – 125 m3 + 150 m2 n + 150 m2 n + 60 m n2 – 60 m n2 + 8 n3 + 8 n3

= 300 m2 n + 16 n3

Therefore, (5m + 2n)3 – (5m – 2n)3 = 300 m2 n + 16 n3

Note- “Cubing” is the procedure of raising a value to the power of 3, which means to multiply a value three times, like- 103 = 10 x 10 x10.

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