Question

Find the cube of the following binomials expression :

\[\frac{1}{x} + \frac{y}{3}\]

Answer 1

In the given problem, we have to find cube of the binomial expressions

Given `(1/x+y/3)^3`

We shall use the identity  `(a+b)^3 = a^3+b^3+3ab(a+b)`

Here  `a=1/x ,b=y/3`

By applying the identity we get 

`(1/x+y/x)^3 = (1/x)^3 + (y/3)^3+3 (1/x)(y/3)(1/x+y/3)`

                      ` = 1/x^3 +y^3/27+3 xx 1/x xx y/3 (1/x +y/3)`

                     ` = 1/x^3 +y^3/27+ y/x (1/x +y/3)`

                     ` = 1/x^3 +y^3/27+ y/x xx 1/x +y/x xxy/3`

                     ` = 1/x^3 +y^3/27+ y/x^2+y/(3x)`

Hence cube of the binomial expression  `1/x+y/3`  is  `1/x^3 + y^3/27 +y/x^2+y^2/(3x)`