<s>[INST] 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
 [/INST] Let us utilize the fact that $a,b,c$ are real
Using this, 
either $\displaystyle(i) ab+bc+ca=0$ 
$\displaystyle\implies bc-a^2=-(ab+ca)-a^2=-a(a+b+c)$
$\displaystyle\implies \frac a{(bc-a^2)^2}=\frac a{a^2(a+b+c)^2}=\frac{bc}{abc(a+b+c)^2} $
$\displaystyle\implies \sum_{\text{cyc}}\frac a{(bc-a^2)^2}=\frac{bc+ca+ab}{abc(a+b+c)^2}=\cdots $
or $\displaystyle(ii) a^2+b^2+c^2-ab-bc-ca=0$
$\displaystyle\implies (a-b)^2+(b-c)^2+(c-a)^2=0$
Now the sum of square of three real numbers is zero, so each individually must be equal to zero
Then  $\displaystyle a=b=c\implies bc-a^2=0$ which is impossible
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