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)`



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