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:

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