Question

Find the cube of the following binomials expression :

\[\frac{3}{x} - \frac{2}{x^2}\]

Answer 1

Given  `(3/x-2/x^2)^3`

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

Here `a=3/x,b = 2/x^2`

By applying the identity we get 

 `(3/x-2/x^2)^3 = (3/x)^3 - (2/x^2)^3 -3 (3/x)(2/x^2)(3/x-2/x^2)`

                       `= 27/x^3 - 8/x^6 -3 xx3/x xx 2/x^2 (3/x - 2/x^2)`

                      `= 27/x^3 - 8/x^6 -18/x^3(3/x - 2/x^2)`

                     `= 27/x^3 - 8/x^6 -(18/x^3 xx3/x) -(18/x^3xx2/x^2)`

                    `= 27/x^3 - 8/x^6 -(54/x^4 +36/x^5)`

                    `= 27/x^3 - 8/x^6 -54/x^4 +36/x^5`

Hence cube of the binomial expression of  `(3/x-2/x^2)` is `= 27/x^3 - 8/x^6 -54/x^4 +36/x^5`.