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.
 A: Assuming
$$\frac a{b-c}+\frac b{c-a}+\frac c{a-b}=0$$
we also have
$$\frac a{(b-c)^2}+\frac b{(c-a)(b-c)}+\frac c{(a-b)(b-c)}=0$$
as well as
$$\frac a{(b-c)(a-b)}+\frac b{(c-a)(a-b)}+\frac c{(a-b)^2}=0$$
and
$$\frac a{(b-c)(c-a)}+\frac b{(c-a)^2}+\frac c{(a-b)(c-a)}=0$$
These three sum together as
$$\frac a{(b-c)^2}+\frac b{(c-a)^2}+\frac c{(a-b)^2}\\+\frac {a+c}{(a-b)(b-c)}+\frac {a+b}{(c-a)(b-c)}+\frac{b+c}{(c-a)(a-b)}=0\tag 1$$
With the three later fractions, we can multiply by $1$ as follows:
$$\frac {(a+c)(c-a)}{(a-b)(b-c)(c-a)}=\frac {c^2-a^2}{(a-b)(b-c)(c-a)}$$
Doing this to each yields
$$\frac {c^2-a^2}{(a-b)(b-c)(c-a)}+\frac {b^2-c^2}{(a-b)(b-c)(c-a)}+\frac {a^2-b^2}{(a-b)(b-c)(c-a)}=0$$
This equivalence should be clear by inspection, which transforms $(1)$ into
$$\frac a{(b-c)^2}+\frac b{(c-a)^2}+\frac c{(a-b)^2} = 0$$
A: 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.
A: Let 
$$
T_1=\sum_{cyc} \frac{a}{b-c}
$$
By hypothesis we have $T_1=0$, but if we put
$T_2=T_1(ab+ac+bc-a^2-b^2-c^2)$, we have 
$$
T_2=\sum_{cyc} \frac{a(ab+ac+bc-a^2-b^2-c^2)}{b-c}
$$
$$
T_2=\sum_{cyc} \frac{a^2b}{b-c}+
\sum_{cyc} \frac{a^2c}{b-c}+
\sum_{cyc} \frac{abc}{b-c}-
\sum_{cyc} \frac{a^3}{b-c}-
\sum_{cyc} \frac{ab^2}{b-c}-
\sum_{cyc} \frac{ac^2}{b-c}
$$
$$
T_2=\sum_{cyc} \frac{a^2b}{b-c}+
\sum_{cyc} \frac{a^2b}{c-b}+
\sum_{cyc} \frac{abc}{b-c}-
\sum_{cyc} \frac{a^3}{b-c}-
\sum_{cyc} \frac{ab^2}{b-c}-
\sum_{cyc} \frac{ab^2}{c-b}
$$
$$
T_2=
\sum_{cyc} \frac{abc}{b-c}-
\sum_{cyc} \frac{a^3}{b-c}
=\sum_{cyc} \frac{a(bc-a^2)}{b-c}
$$
$$
T_2=\sum_{cyc} \frac{a(bc-a^2)}{b-c}
+\sum_{cyc} \frac{-a^2b}{b-c}
+\sum_{cyc} \frac{-a^2c}{b-c}
$$
and hence
$$
T_2=\sum_{cyc} \frac{a(b-a)(c-a)}{b-c}=
-\sum_{cyc} \frac{a(a-b)(c-a)}{b-c}
$$
So if we put $T_3=\frac{T_2}{(a-b)(b-c)(c-a)}$, we see that
$$
T_3=-\sum_{cyc} \frac{a}{(b-c)^2}
$$
and we are done.
