Prove that $(x + y)^{4} = x^{4} + 4 x^{3} y + 6 x^{2} y^{2} + 4 x y^{3} + y^{4}$ .

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Solution

$(x + y)^{4} = x^{4} + 4 x^{3} y + 6 x^{2} y^{2} + 4 x y^{3} + y^{4}$
First take L.H.S $(x + y)^{4}$
So, the above expression is written as $((x + y)^{2})^{2}$
We know that $(x + y)^{2} = x^{2} + 2 x y + y^{2}, (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x$
Therefore, $(x^{2} + 2 x y + y^{2})^{2}$
Here, $x = x^{2}, y = 2 x y, z = y^{2}$
So, $(x^{2} + 2 x y + y^{2})^{2} = x^{4} + 4 x^{2} y^{2} + y^{4} + 4 x^{3} y + 4 x y^{3} + 2 y^{2} x^{2}$
$=$ $x^{4} + 4 x^{3} y + 6 x^{2} y^{2} + 4 x y^{3} + y^{4}$
L.H.S $=$ R.H.S
$x^{4} + 4 x^{3} y + 6 x^{2} y^{2} + 4 x y^{3} + y^{4} = x^{4} + 4 x^{3} y + 6 x^{2} y^{2} + 4 x y^{3} + y^{4}$
Hence $(x + y)^{4} = x^{4} + 4 x^{3} y + 6 x^{2} y^{2} + 4 x y^{3} + y^{4}$ is proved.

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