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Question

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

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Solution

## To multiply, we will use distributive law as follows: $\left(\frac{1}{3}x-\frac{{y}^{2}}{5}\right)\left(\frac{1}{3}x+\frac{{y}^{2}}{5}\right)\phantom{\rule{0ex}{0ex}}=\left[\frac{1}{3}x\left(\frac{1}{3}x+\frac{{y}^{2}}{5}\right)\right]-\left[\frac{{y}^{2}}{5}\left(\frac{1}{3}x+\frac{{y}^{2}}{5}\right)\right]\phantom{\rule{0ex}{0ex}}=\left[\frac{1}{9}{x}^{2}+\frac{x{y}^{2}}{15}\right]-\left[\frac{x{y}^{2}}{15}+\frac{{y}^{4}}{25}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{9}{x}^{2}+\frac{x{y}^{2}}{15}-\frac{x{y}^{2}}{15}-\frac{{y}^{4}}{25}\phantom{\rule{0ex}{0ex}}=\frac{1}{9}{x}^{2}-\frac{{y}^{4}}{25}$ $\therefore$ $\left(\frac{1}{3}x-\frac{{y}^{2}}{5}\right)\left(\frac{1}{3}x+\frac{{y}^{2}}{5}\right)=\frac{1}{9}{x}^{2}-\frac{{y}^{4}}{25}$ Now, we will put x = $-$1 and y = $-$2 on both the sides to verify the result. $\text{LHS =}\left(\frac{1}{3}x-\frac{{y}^{2}}{5}\right)\left(\frac{1}{3}x+\frac{{y}^{2}}{5}\right)\phantom{\rule{0ex}{0ex}}=\left[\frac{1}{3}\left(-1\right)-\frac{{\left(-2\right)}^{2}}{5}\right]\left[\frac{1}{3}\left(-1\right)+\frac{{\left(-2\right)}^{2}}{5}\right]\phantom{\rule{0ex}{0ex}}=\left(-\frac{1}{3}-\frac{4}{5}\right)\left(-\frac{1}{3}+\frac{4}{5}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{-17}{15}\right)\left(\frac{7}{15}\right)\phantom{\rule{0ex}{0ex}}=\frac{-119}{225}$ $\text{RHS}=\frac{1}{9}{x}^{2}-\frac{{y}^{4}}{25}\phantom{\rule{0ex}{0ex}}=\frac{1}{9}{\left(-1\right)}^{2}-\frac{{\left(-2\right)}^{4}}{25}\phantom{\rule{0ex}{0ex}}=\frac{1}{9}×1-\frac{16}{25}\phantom{\rule{0ex}{0ex}}=\frac{1}{9}-\frac{16}{25}\phantom{\rule{0ex}{0ex}}=-\frac{119}{225}$ Because LHS is equal to RHS, the result is verified. Thus, the answer is $\frac{1}{9}{x}^{2}-\frac{{y}^{4}}{25}$.

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