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) $

Given: 

$x = -2$ and $y = 1$

To do: 

We have to find the value of \( \left(\frac{2}{x}-\frac{x}{2}\right)\left(\frac{4}{x^{2}}+\frac{x^{2}}{4}+1\right) \).

Solution: 

We know that,

$a^{3}+b^{3}=(a+b)(a^{2}-a b+b^{2})$

$a^{3}-b^{3}=(a-b)(a^{2}+a b+b^{2})$

Therefore,

$(\frac{2}{x}-\frac{x}{2})(\frac{4}{x^{2}}+\frac{x^{2}}{4}+1)=(\frac{2}{x}-\frac{x}{2})[(\frac{2}{x})^{2}+2 \times \frac{2}{x} \times \frac{x}{2}+(\frac{x}{2})^{2}]$

$=(\frac{2}{x})^{3}-(\frac{x}{2})^{3}$

$=\frac{8}{x^{3}}-\frac{x^{3}}{8}$

$=\frac{8}{(-2)^{3}}-\frac{(-2)^{3}}{8}$

$=\frac{8}{-8}-\frac{-8}{8}$

$=-1+1$

$=0$

Hence, $(\frac{2}{x}-\frac{x}{2})(\frac{4}{x^{2}}+\frac{x^{2}}{4}+1)=0$.   

Tutorialspoint

Updated on: 10-Oct-2022

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