Find the following products and verify the result for x = -1, y = -2:\

1.	Find the following products and verify the result for x = -1, y = -2:\((\frac{1}{3}x-\frac{y^2}{5})(\frac{1}{3}x+\frac{y^2}{5})\)
Answer» \((\frac{1}{3}x)^2-(\frac{y\times y}{5})^2\) = \((\frac{1}{3}x-\frac{y\times y}{5})(\frac{1}{3}x+\frac{y\times y}{5})\) = \(\frac{1}{9}x^2-\frac{1}{25}y^4\) Putting x = -1 and y = -2, we have \((\frac{1}{3}(-1)-\frac{(-2)(-25)}{5})\) = \((\frac{1}{9}(-1)^2-\frac{-2\times -2\times -2\times -2}{25})\) = \((\frac{-1}{3}-\frac{4}{5})(\frac{-1}{3}+\frac{4}{5})=(\frac{1}{9}-\frac{16}{25})\) = \((\frac{-17}{15})(\frac{7}{15})=\frac{-119}{225}\) = \(\frac{-119}{225}\) = \(\frac{-119}{225}\) Therefore, L.H.S = R.H.S Hence, verified

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

Answer»

\((\frac{1}{3}x)^2-(\frac{y\times y}{5})^2\)

= \((\frac{1}{3}x-\frac{y\times y}{5})(\frac{1}{3}x+\frac{y\times y}{5})\)

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

Putting x = -1 and y = -2, we have

\((\frac{1}{3}(-1)-\frac{(-2)(-25)}{5})\) = \((\frac{1}{9}(-1)^2-\frac{-2\times -2\times -2\times -2}{25})\)

= \((\frac{-1}{3}-\frac{4}{5})(\frac{-1}{3}+\frac{4}{5})=(\frac{1}{9}-\frac{16}{25})\)

= \((\frac{-17}{15})(\frac{7}{15})=\frac{-119}{225}\)

= \(\frac{-119}{225}\) = \(\frac{-119}{225}\)

Therefore,

L.H.S = R.H.S

Hence, verified