RD Sharma Chapter 4 Class 9 Maths Exercise 4.4 Solutions is about the sum and difference of cubes. The algebraic expressions will be used but in the cubical form. With the help of this article, students will learn about how to simplify products of the algebraic expressions, that is, to find out the sums and differences of cubes based on the given problems.

We have attached the RD Sharma Chapter 4 Class 9 Maths Exercise 4.4 Solutions PDF in which several types of questions will be there with the stepwise solutions given by our experts. The explanations of each problem are given in very easy to understand, and the shortcuts of every problem are also provided with the solutions.

**Learn about RD Sharma Class 9 Chapter 4- Algebraic Identities**

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## Download RD Sharma Chapter 4 Class 9 Maths Exercise 4.4 Solutions PDF

Solutions for Class 9 Maths Chapter 4 Algebraic Identities Exercise 4.4

## Important Definitions RD Sharma Chapter 4 Class 9 Maths Exercise 4.4 Solutions

Here the question comes- “What is the Sums and Differences of Cubes? So, the answer is-

- A polynomial in form x
^{3}+ y^{3}is called a sum of cubes. - A polynomial in form x
^{3}– y^{3}is called a difference of cubes.

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

- x
^{3}+ y^{3}= (x + y)^{3}– 3xy (x + y)

**or** x^{3} + y^{3} = (x + y) (x^{2} – xy + y^{2})

- x
^{3}– y^{3}= (x – y)^{3}– 3xy (x – y)

**or** x^{3} – y^{3} = (x – y) (x^{2} + xy + y^{2})

Go down to check some examples of the Sums and Differences of Cubes with algebraic expressions-

**Ques 1- If p + q = 8 and pq = 6, find the value of p ^{3} + q^{3}.**

**Solution-**

p+q = 8, pq = 6

Cubing, p+q = 8, both sides, we get

(p + q)^{3} = (8)^{3}

p^{3} + q^{3} + 3pq (p + q) = 512

p^{3} + q^{3} + 3 x 6 x 8 = 512

p^{3} + q^{3} + 144 = 512

p^{3} + q^{3} = 512 – 144 = 368

p^{3} + q^{3} = 368

**Ques 2- If x + y = 8 and xy = 6, find the value of x ^{3} + y^{3}.**

**Solution:**

x + y = 8, xy = 6

Cubing both sides,

(x + y)^{3} = (8)^{3}

= x^{3} + y^{3} + 3xy (x + y) = 512

= x^{3} + y^{3} + 3 x 6 x 8 = 512

= x^{3} + y^{3} + 144 = 512

= x^{3} + y^{3} = 512 – 144 = 368

Therefore, x^{3} + y^{3} = 368

**Ques 3- If a – b = 6 and ab = 20, find the value of a ^{3}– b^{3}.**

**Solution:**

p – q = 6, pq = 20

Cubing both sides,

(p – q)3 = (6)^{3}

= p^{3} – q^{3} – 3pq (p – q) = 216

= p^{3} – q^{3} – 3 x 20 x 6 = 216

= p^{3} – q^{3} – 360 = 216

= p^{3} – q^{3} = 216 + 360 = 576

Therefore, p^{3} – q^{3} = 576

## Frequently Asked Questions (FAQs) of RD Sharma Chapter 4 Class 9 Maths Exercise 4.4 Solutions

**Ques 1- How do you find the sum of a cube?**

**Ans-** It shows that the sum is constantly square, but what is even more exceptional is that the sum of the first n cubes, 13+23+… + n 3 = (n (n +1)/2)2, which is the square of the nth triangle number.

For Example-

13+23+…+103 = (10×11/2)2 = 552 = 3025.

**Ques 2- How do you factor the sum and difference of two cubes?**

**Ans-** The sum or difference of two cubes will be factored into a product of binomial times a trinomial.