# Find the following products and verify the result for $x = -1, y = -2$:$\left(\frac{1}{3} x-\frac{y^{2}}{5}\right)\left(\frac{1}{3} x+\frac{y^{2}}{5}\right)$

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Given:

$\left(\frac{1}{3} x-\frac{y^{2}}{5}\right)\left(\frac{1}{3} x+\frac{y^{2}}{5}\right)$

To do:

We have to find the given product and verify the result for $x = -1, y = -2$.

Solution:

We know that,

$(a+b)\times(c+d)=a(c+d)+b(c+d)$

Therefore,

$(\frac{1}{3} x-\frac{y^{2}}{5})(\frac{1}{3} x+\frac{y^{2}}{5})=[\frac{1}{3} x(\frac{1}{3} x+\frac{y^{2}}{5})]-[\frac{y^{2}}{5}(\frac{1}{3} x+\frac{y^{2}}{5})]$

$=[\frac{1}{9} x^{2}+\frac{x y^{2}}{15}]-[\frac{x y^{2}}{15}+\frac{y^{4}}{25}]$

$=\frac{1}{9} x^{2}+\frac{x y^{2}}{15}-\frac{x y^{2}}{15}-\frac{y^{4}}{25}$

$=\frac{1}{9} x^{2}-\frac{y^{4}}{25}$

LHS $=(\frac{1}{3} x-\frac{y^{2}}{5})(\frac{1}{3} x+\frac{y^{2}}{5})$

$=[\frac{1}{3}(-1)-\frac{(-2)^{2}}{5}][\frac{1}{3}(-1)+\frac{(-2)^{2}}{5}]$

$=(-\frac{1}{3}-\frac{4}{5})(-\frac{1}{3}+\frac{4}{5})$

$=(\frac{-17}{15})(\frac{7}{15})$

$=\frac{-119}{225}$

RHS $=\frac{1}{9} x^{2}-\frac{y^{4}}{25}$

$=\frac{1}{9}(-1)^{2}-\frac{(-2)^{4}}{25}$

$=\frac{1}{9} \times 1-\frac{16}{25}$

$=\frac{1}{9}-\frac{16}{25}$

$=-\frac{119}{225}$

Therefore,

LHS $=$ RHS

Updated on 10-Oct-2022 13:19:30