Show these two equations for variance are equivalent This is probably not too difficult, but I am having a lot of trouble with it.
I am trying to show the following is true:
$${\sum (x_i - \bar x)\over n-1} = {\sum x_i^2 - {(\sum x_i)^2\over n}\over n - 1}$$
$\bar x$ is the mean of $x_i$ or
$$\sum x \over n$$
I have made some progress but cannot solve it myself. Thank you for any help.

I'm assuming you on the left hand side want $\sum (x_i-\bar{x})^2$ instead of $\sum\left(x_i-\bar{x}\right)$ because the latter is just $0$. Just expand the square and we obtain
$$
\begin{align}
\sum (x_i-\bar{x})^2&=\sum \left(x_i^2+\bar{x}^2-2x_i\bar{x}\right)=\sum x_i^2+n\bar{x}^2-2\bar{x}\sum x_i\\
&=\sum x_i^2+n\bar{x}^2-2n\bar{x}^2=\sum x_i^2-n\bar{x}^2=\sum x_i^2-\frac{\left(\sum x_i\right)^2}{n}.
\end{align}
$$
