How prove this $\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}=0$ let $a,b,c\in \mathbb{R}$, if such

$$\dfrac{1}{bc-a^2}+\dfrac{1}{ca-b^2}+\dfrac{1}{ab-c^2}=0$$
show that
$$\dfrac{a}{(bc-a^2)^2}+\dfrac{b}{(ca-b^2)^2}+\dfrac{c}{(ab-c^2)^2}=0$$

Does this problem has nice methods?
My idea:let
$$(ca-b^2)(ab-c^2)+(bc-a^2)(ab-c^2)+(bc-a^2)(ca-b^2)=0$$
then I can't. Thanks

Hint: if we have $$\dfrac{1}{bc-a^2}+\dfrac{1}{ca-b^2}+\dfrac{1}{ab-c^2}=0$$ Then let $$\dfrac{1}{bc-a^2} = x; \dfrac{1}{ca-b^2} =y; \dfrac{1}{ab-c^2} = z$$ So you have something more manageable, then note that if $x+y+z=0$, then $x = -(y+z)$. 
Finally, note that $$\dfrac{a}{(bc-a^2)^2} = \dfrac{a}{bc-a^2}\cdot \dfrac{1}{bc-a^2} = \dfrac{a}{bc-a^2}\cdot x$$
Etc. From here, you should be able to make progress.
