If $x = 3$ and $y = -1$, find the values of each of the following using in identity:$ \left(\frac{x}{4}-\frac{y}{3}\right)\left(\frac{x^{2}}{16}+\frac{x y}{12}+\frac{y^{2}}{9}\right) $

Given: 

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

To do: 

We have to find the value of \( \left(\frac{x}{4}-\frac{y}{3}\right)\left(\frac{x^{2}}{16}+\frac{x y}{12}+\frac{y^{2}}{9}\right) \).

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,

$(\frac{x}{4}-\frac{y}{3})(\frac{x^{2}}{16}+\frac{x y}{12}+\frac{y^{2}}{9})=(\frac{x}{4}-\frac{y}{3})[(\frac{x}{4})^{2}+\frac{x}{4} \times \frac{y}{3}+(\frac{y}{3})^{2}]$

$=(\frac{x}{4})^{3}-(\frac{y}{3})^{3}$

$=\frac{x^{3}}{64}-\frac{y^{3}}{27}$

$=\frac{(3)^{3}}{64}-\frac{(-1)^{3}}{27}$

$=\frac{27}{64}+\frac{1}{27}$

$=\frac{729+64}{1728}$

$=\frac{793}{1728}$

Hence, $(\frac{x}{4}-\frac{y}{3})(\frac{x^{2}}{16}+\frac{x y}{12}+\frac{y^{2}}{9})=\frac{793}{1728}$.

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Updated on: 10-Oct-2022

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