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.

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
$$\frac a{(b-c)^2}+\frac b{(c-a)^2}+\frac c{(a-b)^2} = 0$$
