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:

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

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

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

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

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

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

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

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

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

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

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

\[ = \frac{- 119}{225}\]

\[\text { RHS } = \frac{1}{9} x^2 - \frac{y^4}{25}\]

\[ = \frac{1}{9} \left( - 1 \right)^2 - \frac{\left( - 2 \right)^4}{25}\]

\[ = \frac{1}{9} \times 1 - \frac{16}{25}\]

\[ = \frac{1}{9} - \frac{16}{25}\]

\[ = - \frac{119}{225}\]

Because LHS is equal to RHS, the result is verified.
Thus, the answer is \[\frac{1}{9} x^2 - \frac{y^4}{25}\].