Find the following product and verify the result for x = − 1, y = − 2:

Solution:

To multiply, we will use distributive law as follows:

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

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

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

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

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

$\therefore\left(\frac{1}{3} x-\frac{y^{2}}{5}\right)\left(\frac{1}{3} x+\frac{y^{2}}{5}\right)=\frac{1}{9} x^{2}-\frac{y^{4}}{25}$

Now, we will put $x=-1$ and $y=-2$ on both the sides to verify the result.

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

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

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

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

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

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

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

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

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

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

Because LHS is equal to RHS, the result is verified.

Thus, the answer is $\frac{1}{9} x^{2}-\frac{y^{4}}{25}$.