Problems with understanding a given Sum Identity In a textbook I found the following equation without any explaination:

\begin{align}
\sum_{\substack{j, k=1,\\ j\neq k}}^n\frac{1}{(x-x_j)(x-x_k)} - \frac{3}{2}\left( \sum_{j=1}^n \frac{1}{x-x_j}\right)^2 \\ 
 = -\frac{1}{2}\sum_{j=1}^n \left( \frac{1}{x-x_j}\right)^2 - \left( \sum_{j=1}^n \frac{1}{x-x_j} \right)^2
\end{align}

I was wondering why this is true and tried to proof it but somehow failed. Can anyone give a hint?
There is a misrpint here. The factor $\frac 1 2$ should apply to the second term on the right side, not the first.

This modified identity follows immediately from the fact that the first term on the left side is $ \sum\limits_{j,k=1}^{n} \frac 1 {(x-x_j)(x-x_k)}-\sum\limits_{k=1}^{n} \frac 1 {(x-x_j)^{2}}$ which is same as $ (\sum\limits_{k=1}^{n} \frac 1 {(x-x_j)})^{2}-\sum\limits_{k=1}^{n} \frac 1 {(x-x_j)^{2}}$
Consider the following expression
\begin{eqnarray*}
\left( \sum_{j=1}^{n} \frac{1}{x-x_j} \right)^2 = \left( \sum_{j=1}^{n} \frac{1}{x-x_j} \right) \left( \sum_{k=1}^{n} \frac{1}{x-x_k} \right).
\end{eqnarray*}
Now when you calculate the product we have either $j \neq k$ or $j=k$, so we have 
\begin{eqnarray*}
\left( \sum_{j=1}^{n} \frac{1}{x-x_j} \right)^2 =  \sum_{\substack{j, k=1,\\ j\neq k}}^n \frac{1}{x-x_j} + \sum_{j=1}^{n} \frac{1}{(x-x_j)^2} .
\end{eqnarray*}
So the first term in the stated formula needs a minus sign!