$(x + y)^{2} = (x - y)^{2} + 4 x y$ prove this identity.

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Solution

$(x + y)^{2} = (x - y)^{2} + 4 x y$
First take R.H.S
$(x - y)^{2} + 4 x y$
We know that $(x - y)^{2} = x^{2} - 2 x y + y^{2}, (x + y)^{2} = x^{2} + 2 x y + y^{2}$
Therefore, $x^{2} - 2 x y + y^{2} + 4 x y$
$=$ $x^{2} + 2 x y + y^{2}$
$=$ $(x + y)^{2}$
So, L.H.S $=$ R.H.S
$(x + y)^{2} = (x + y)^{2}$
Hence $(x + y)^{2} = (x - y)^{2} + 4 x y$ is proved.

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