Answer
$Σ(x_i-x ̅)^2=Σx_i^2-\frac{1}{n}(Σx_i)^2$
Work Step by Step
$Σ(x_i-x ̅)^2=Σ(x_i^2-2x_ix ̅+x ̅^2)=Σx_i^2-2x ̅Σx_i+x ̅^2Σ(1)$
But, $Σ(1)=1+1+1+...+1=n$
$\frac{1}{n}Σx_i=x ̅$. So:
$\frac{1}{n^2}(Σx_i)^2=x ̅^2$
$Σ(x_i-x ̅)^2=Σx_i^2-2x ̅nx ̅+nx ̅^2=Σx_i^2-nx ̅^2=Σx_i^2-n\frac{1}{n^2}(Σx_i)^2=Σx_i^2-\frac{1}{n}(Σx_i)^2$
