Question

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

\[\left( \frac{2}{x} - \frac{x}{2} \right) \left( \frac{4}{x^2} + \frac{x^2}{4} + 1 \right)\]

Answer 1

In the given problem, we have to find the value of \[\left( \frac{2}{x} - \frac{x}{2} \right) \left( \frac{4}{x^2} + \frac{x^2}{4} + 1 \right)\] using identity

Given x = -2

We shall use the identity  `(a+b)(a^2 + ab + b^2) = (a^3 - b^3)`

We can rearrange the  \[\left( \frac{2}{x} - \frac{x}{2} \right) \left( \frac{4}{x^2} + \frac{x^2}{4} + 1 \right)\]as

`(2/x - x/2)(4/x^2 + x^2/4 + 1) = (2/x - x/2)((2/x)^2 +2/x xx x/2+ (x/2)^2)`

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

\[= \left( \frac{2}{x} \right) \times \left( \frac{2}{x} \right) \times \left( \frac{2}{x} \right) - \left( \frac{x}{2} \right) \times \left( \frac{x}{2} \right) \times \left( \frac{x}{2} \right)\]
\[ = \frac{8}{x^3} - \frac{x^3}{8}\]

Now substituting the value  x = -2 in  `8/x^2 - x^3/8`we get,

` = 8/(-2)^3  - ( -2)^3/8`

` = 8/-8 - -8/8`

 ` =  -1 - (-1)`

` = -1+1`

` = 0`

Hence the Product value of  \[\left( \frac{2}{x} - \frac{x}{2} \right) \left( \frac{4}{x^2} + \frac{x^2}{4} + 1 \right)\] is = 0.