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?

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 $(bc-a^2)(ab-c^2)(ca-b^2)\ne0$
${\quad a,b,c\in \mathbb{R}, {\text { is not actually required}}}$
