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) $
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
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