Question

Show that $x^2+y^2=\frac{1}{2}\left[\right(x+y{)}^2+(x-y{)}^2]$.

Answer 1

$x^2+y^2=\frac{1}{2}\left[\right(x+y{)}^2+(x-y{)}^2]$
First take R.H.S
$\frac{1}{2}\left[\right(x+y{)}^2+(x-y{)}^2]$
We know that $(x+y{)}^2=x^2+2xy+y^2,(x-y{)}^2=x^2-2xy+y^2$
Therefore, $\frac{1}{2}[x^2+2xy+y^2+x^2-2xy+y^2]$
$=$ $\frac{1}{2}[2x^2+2y^2]$
$=$ $\frac{1}{2}\left[2\right(x^2+y^2\left)\right]$ (cancelling $2$)
$=$ $x^2+y^2$
So, L.H.S = R.H.S
$x^2+y^2=x^2+y^2$
Hence $x^2+y^2=(x+y{)}^2-2xy$ is proved.