# Polynomials

## Exercise 2.5 Part 7

Question: 6 – Write the following cubes in expanded form:

(i) (2x + 1)^3

Answer: Given, (2x + 1)^3

Let, a = 2x and b = 1

Using the identity (a + b)^3= a^3 + b^3 + 3ab (a + b)

We get;
=(2x)^3 + 1^3 + 3xx(2x)xx1(2x + 1)

= 8x^3 + 1 + 12x^2 + 6x

= 8x^3 + 12x^2 + 6x + 1

(ii) (2a – 3b)^3

Answer: Given; (2a – 3b)^3
Let, x = 2a and y = 3b

Using the identity (x – y)^3= x^3 - y^3 - 3yx^2 + 3xy^2

The given expression can be written as
(2a)^3 - (3b)^3 - 3(3b)(2a)^2 + 3(2a)(3b)^2

= 8a^3 - 27b^3 - (9b) xx 4a^2 + (6a) xx 9b^2

= 8a^3 - 27b^3 - 36a^2b + 54ab^2

(iii) (3/2\x+1)^3

Solution: Given, (3/2\x+1)^3

Let, a=3/2\x and b=1

[Using identity (a+b)^3=a^3+b^3+3ab(a+b) we get

(3/2\x)^3+1^3+3xx(3/2\x)xx1(3/2\x+1)

=27/8\x^3+1+9/2\x(3/2\x+1)

=(27x^3)/(8)+1+(27x^2)/(4)+(9x)/(2)

=(27x^3)/(8) +(27x^2)/(4)+(9x)/(2) +1

(iv) (x-2/3\y)^3

Solution: Given (x-(2)/(3)\y)^3

Let a=x and b=2/3\y

Using identity (x-y)^3=x^3-y^3-3x^2y+3xy^2

Given expression can be written as follows:

x^3-(2/3\y)^3-3x^2(2/3\y)+3x(2/3\y)^2

=x^3-(8y^3)/(27)-2x^2y+(3x)xx(4y^2)/(9)

= x^3-(8y^3)/(27)-2x^2y+(4xy^2)/(3)