Find a cyclic sum Given that
$$ \frac{a}{bc-a^2}+\frac{b}{ca-b^2}+\frac{c}{ab-c^2}=0,\quad a,b,c\in \mathbb{R}, $$
find the value of
$$ \frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}. $$
I solved it via direct computation which is clearly not the best way to do. Is there any simple transformation I could perform or geometical meaning of this equation?
 A: From first equation (get rid of the denumerators) we get $$abc(a^2+b^2+c^2+ab+bc+ca)= a^2b^2(a+b)+b^2c^2(b+c)+c^2a^2(c+a)$$ which can be easly transforemd to $$abc(a^2+b^2+c^2+2ab+2bc+2ca)= (a^2b^2+b^2c^2+c^2a^2)(a+b+c)$$ so $$abc(a+b+c)^2= (a^2b^2+b^2c^2+c^2a^2)(a+b+c)$$
Now if $a+b+c\neq 0$ then we have $$\color{red}{abc(a+b+c)= a^2b^2+b^2c^2+c^2a^2}$$
But since \begin{align}a^2b^2 +b^2c^2&\geq 2b^2ac \\
b^2c^2 +c^2a^2&\geq 2c^2ab \\
c^2a^2 +a^2b^2&\geq 2a^2bc \\
\end{align}
we have always $$abc(a+b+c)\leq  a^2b^2+b^2c^2+c^2a^2$$
with equality iff $a=b=c$. So, since we have equality sign in red equation, we deduce that $a=b=c$, but this can not hold since then fractions are not defined.
So $\boxed{a+b+c =0}$ and now I suppose it is not difficult to finish....


 Let $E$ be the expression. Then $$E = b\Big(\frac{1}{(ca-b^2)^2}-\frac{1}{(bc-a^2)^2}\Big)+      c\Big(\frac{1}{(ab-c^2)^2} -\frac{1}{(bc-a^2)^2}\Big) $$ Value in each big bracket is $0$ so $E=0$.

A: Hint:
$$\left(\dfrac a{bc-a^2}+\dfrac b{ca-b^2}+\dfrac c{ab-c^2}\right)\cdot{\underbrace{\left(\dfrac1{bc-a^2}+\dfrac1{ca-b^2}+\dfrac1{ab-c^2}\right)}_{\text{assuming this to be finite}}}$$
$$= \underbrace{\dfrac a{(bc-a^2)^2}+\dfrac b{(ca-b^2)^2}+\dfrac c{(ab-c^2)^2}}_{\text{this is to be calculated}}+\dfrac{(a+b)(ab-c^2)+(b+c)(bc-a^2)+(c+a)(ca-b^2)}{(bc-a^2)(ab-c^2)(ca-b^2)}$$
Now $(a+b)(ab-c^2)+(b+c)(bc-a^2)+(c+a)(ca-b^2)=\cdots=0$
We only need $(bc-a^2)(ab-c^2)(ca-b^2)\ne0$
${\quad a,b,c\in \mathbb{R}, {\text { is not actually required}}}$
