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

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.4

Question 1.

Find the following products:

(i) (3x + 2y) (9X^{2} – 6xy + Ay^{2})

(ii) (4x – 5y) (16x^{2} + 20xy + 25y^{2})

(iii) (7p^{4} + q) (49p^{8} – 7p^{4}q + q^{2})

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 a^{2} – ab + b^{2} and a^{2} + ab + b^{2}.

Solution:

a + b = 10, ab = 16 Squaring,

(a + b)^{2} = (10)^{2
}⇒ a^{2} + b^{2} + lab = 100

⇒ a^{2} + b^{2} + 2 x 16 = 100

⇒ a^{2} + b^{2} + 32 = 100

∴ a^{2} + b^{2} = 100 – 32 = 68

Now, a^{2} – ab + b^{2} = a^{2} + b^{2} – ab = 68 – 16 = 52

and a^{2} + ab + b^{2} = a^{2} + b^{2} + ab = 68 + 16 = 84

Question 4.

If a + b = 8 and ab = 6, find the value of a^{3} + b^{3}.

Solution:

a + b = 8, ab = 6

Cubing both sides,

(a + b)^{3} = (8)3

⇒ a^{3} + b^{3} + 3 ab{a + b) = 512

⇒ a^{3} + b^{3} + 3 x 6 x 8 = 512_{
}⇒ a^{3} + b^{3} + 144 = 512

⇒ a^{3} + b^{3} = 512 – 144 = 368

∴ a^{3} + b^{3} = 368

Question 5.

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

Solution:

a – b = 6, ab = 20

Cubing both sides,

(a – b)3 = (6)^{3}

⇒ a^{3} – b^{3} – 3ab(a – b) = 216

⇒ a^{3} – b^{3} – 3 x 20 x 6 = 216

⇒ a^{3} – b^{3} – 360 = 216

⇒ a^{3} -b^{3} = 216 + 360 = 576

∴ a^{3} – b^{3} = 576

Question 6.

If x = -2 and y = 1, by using an identity find the value of the following:

Solution:

Hope given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.4 are helpful to complete your math homework.

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