If $x = 3$ and $y = -1$, find the values of each of the following using in identity:$ \left(\frac{x}{7}+\frac{y}{3}\right)\left(\frac{x^{2}}{49}+\frac{y^{2}}{9}-\frac{x y}{21}\right) $

Given: 

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

To do: 

We have to find the value of \( \left(\frac{x}{7}+\frac{y}{3}\right)\left(\frac{x^{2}}{49}+\frac{y^{2}}{9}-\frac{x y}{21}\right) \).

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,

$(\frac{x}{7}+\frac{y}{3})(\frac{x^{2}}{49}+\frac{y^{2}}{9}-\frac{x y}{21})=(\frac{x}{7}+\frac{y}{3})[(\frac{x}{7})^{2}-\frac{x}{7} \times \frac{y}{3}-(\frac{y}{3})^{2}]$

$=(\frac{x}{7})^{3}+(\frac{y}{3})^{3}$

$=\frac{x^{3}}{343}+\frac{y^{3}}{27}$

$=\frac{(3)^{3}}{343}+\frac{(-1)^{3}}{27}$

$=\frac{27}{343}-\frac{1}{27}$

$=\frac{729-343}{9261}$

$=\frac{386}{9261}$

Hence, $(\frac{x}{7}+\frac{y}{3})(\frac{x^{2}}{49}+\frac{y^{2}}{9}-\frac{x y}{21})=\frac{386}{9261}$.

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Updated on: 10-Oct-2022

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