If $\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0$ with unequal $a,b,c$, Prove that $\dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)^2}+\dfrac{c}{(a-b)^2}=0$ If $\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0$ with unequal $a,b,c$, Prove that $\dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)^2}+\dfrac{c}{(a-b)^2}=0$  
I could not approach the problem at all though I think I could have done something by using Cauchy-Schwarz inequality, but could not pull this together. Please help.

You only need to prove that
$$
\begin{multline}
\left(\frac{a}{b - c} + \frac{b}{c - a} + \frac{c}{a - b}\right)\left(\frac{1}{b - c} + \frac{1}{c - a} + \frac{1}{a - b}\right) \\
= \frac{a}{(b - c)^2} + \frac{b}{(c - a)^2} + \frac{c}{(a - b)^2}
\end{multline}$$

Where did that come from?
Well, consider this simple equality
$$
\frac{a}{b - c}\left(\color{green}{\frac{1}{c - a} + \frac{1}{a - b}} + \frac{1}{b - c}\right) = \color{green}{\frac{ac - ab}{(a - b)(b - c)(c - a)}} + \frac{a}{(b - c)^2}
$$
When you take cyclic sum of left and right hand sides, the first fraction in RHS dissapears.
In other words, we can write 2 similar ones:
$$
\begin{align}
\frac{b}{c - a}\left(\frac{1}{b - c} + \frac{1}{a - b} + \frac{1}{c - a}\right) = \frac{ab - bc}{(a - b)(b - c)(c - a)} + \frac{b}{(c - a)^2} \\
\frac{c}{a - b}\left(\frac{1}{b - c} + \frac{1}{c - a} + \frac{1}{a - b}\right) = \frac{bc - ac}{(a - b)(b - c)(c - a)} + \frac{c}{(a - b)^2}
\end{align}
$$
Three last equations add up to the first equation since those big fractions right after the equal sign cancel out.
