Find the following products:$ \left(\frac{2}{x}+3 x\right)\left(\frac{4}{x^{2}}+9 x^{2}-6\right) $

Given: 

\( \left(\frac{2}{x}+3 x\right)\left(\frac{4}{x^{2}}+9 x^{2}-6\right) \)

To do: 

We have to find the given product.

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,

$(\frac{2}{x}+3 x)(\frac{4}{x^{2}}+9 x^{2}-6)=(\frac{2}{x}+3 x)(\frac{4}{x^{2}}-6+9 x^{2})$

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

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

$=\frac{8}{x^{3}}+27 x^{3}$

 Hence, $(\frac{2}{x}+3 x)(\frac{4}{x^{2}}+9 x^{2}-6)=\frac{8}{x^{3}}+27 x^{3}$.

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Updated on: 10-Oct-2022

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