Question

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

Answer 1

To multiply, we will use distributive law as follows:

$$\begin{gathered} \left(\frac{1}{3}x-\frac{y^2}{5}\right)\left(\frac{1}{3}x+\frac{y^2}{5}\right) \\ =\left[\frac{1}{3}x\left(\frac{1}{3}x+\frac{y^2}{5}\right)\right]-\left[\frac{y^2}{5}\left(\frac{1}{3}x+\frac{y^2}{5}\right)\right] \\ =\left[\frac{1}{9}{x}^2+\frac{x{y}^2}{15}\right]-\left[\frac{x{y}^2}{15}+\frac{y^4}{25}\right] \\ =\frac{1}{9}{x}^2+\frac{x{y}^2}{15}-\frac{x{y}^2}{15}-\frac{y^4}{25} \\ =\frac{1}{9}{x}^2-\frac{y^4}{25} \end{gathered}$$

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

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

$$\begin{gathered} \text{LHS =}\left(\frac{1}{3}x-\frac{y^2}{5}\right)\left(\frac{1}{3}x+\frac{y^2}{5}\right) \\ =\left[\frac{1}{3}\left(-1\right)-\frac{{\left(-2\right)}^2}{5}\right]\left[\frac{1}{3}\left(-1\right)+\frac{{\left(-2\right)}^2}{5}\right] \\ =\left(-\frac{1}{3}-\frac{4}{5}\right)\left(-\frac{1}{3}+\frac{4}{5}\right) \\ =\left(\frac{-17}{15}\right)\left(\frac{7}{15}\right) \\ =\frac{-119}{225} \end{gathered}$$

$$\begin{gathered} \text{RHS}=\frac{1}{9}{x}^2-\frac{y^4}{25} \\ =\frac{1}{9}{\left(-1\right)}^2-\frac{{\left(-2\right)}^4}{25} \\ =\frac{1}{9}\times 1-\frac{16}{25} \\ =\frac{1}{9}-\frac{16}{25} \\ =-\frac{119}{225} \end{gathered}$$

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

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