# RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.4

In this chapter, we provide RD Sharma Solutions for Chapter 4 Algebraic Identities Ex 4.4 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Chapter 4 Algebraic Identities Ex 4.4 Maths pdf, free RD Sharma Solutions for Chapter 4 Algebraic Identities Ex 4.4 Maths book pdf download. Now you will get step by step solution to each question.

### RD Sharma Solutions for Class 9 Chapter 4 Algebraic Identities Ex 4.4 Download PDF

Question 1.
Find the following products:
(i) (3x + 2y) (9X2 – 6xy + Ay2)
(ii) (4x – 5y) (16x2 + 20xy + 25y2)
(iii) (7p4 + q) (49p8 – 7p4q + q2) Solution:    Question 2.
If x = 3 and y = -1, find the values of each of the following using in identity: Solution:    Question 3.
If a + b = 10 and ab = 16, find the value of a2 – ab + b2 and a2 + ab + b2.
Solution:
a + b = 10, ab = 16 Squaring,
(a + b)2 = (10)2
⇒ a2 + b2 + lab = 100
⇒ a2 + b2 + 2 x 16 = 100
⇒  a2 + b2 + 32 = 100
∴ a2 + b2 = 100 – 32 = 68
Now, a2 – ab + b2 = a2 + b2 – ab = 68 – 16 = 52
and a2 + ab + b2 = a2 + b2 + ab = 68 + 16 = 84

Question 4.
If a + b = 8 and ab = 6, find the value of a3 + b3.
Solution:
a + b = 8, ab = 6
Cubing both sides,
(a + b)3 = (8)3
⇒ a3 + b3 + 3 ab{a + b) = 512
⇒  a3 + b3 + 3 x 6 x 8 = 512⇒  a3 + b3 + 144 = 512
⇒  a3 + b3 = 512 – 144 = 368
∴ a3 + b3 = 368

Question 5.
If a – b = 6 and ab = 20, find the value of a3-b3.
Solution:
a – b = 6, ab = 20
Cubing both sides,
(a – b)3 = (6)3
⇒  a3 – b3 – 3ab(a – b) = 216
⇒  a3 – b3 – 3 x 20 x 6 = 216
⇒  a3 – b3 – 360 = 216
⇒  a3 -b3 = 216 + 360 = 576
∴ a3 – b3 = 576

Question 6.
If x = -2 and y = 1, by using an identity find the value of the following: Solution: All Chapter RD Sharma Solutions For Class 9 Maths

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All Subject NCERT Exemplar Problems Solutions For Class 9

All Subject NCERT Solutions For Class 9

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