If $x = 3$ and $y = -1$, find the values of each of the following using in identity:$ \left(\frac{3}{x}-\frac{x}{3}\right)\left(\frac{x^{2}}{9}+\frac{9}{x^{2}}+1\right) $

Given: 

$x = 3$ and $y = -1$

To do: 

We have to find the value of \( \left(\frac{3}{x}-\frac{x}{3}\right)\left(\frac{x^{2}}{9}+\frac{9}{x^{2}}+1\right) \).

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,

$(\frac{3}{x}-\frac{x}{3})(\frac{x^{2}}{9}+\frac{9}{x^{2}}+1)=(\frac{3}{x}-\frac{x}{3})[(\frac{x}{3})^{2}+\frac{x}{3} \times \frac{3}{x}+(\frac{3}{x})^{2}]$

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

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

$=1^{3}-1^{3}$

$=0$

Hence, $(\frac{3}{x}-\frac{x}{3})(\frac{x^{2}}{9}+\frac{9}{x^{2}}+1)=0$. 

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Updated on: 10-Oct-2022

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