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

One approach uses factoring. Define rational functions
$$ R_1 := \frac1{b c-a a} + \frac1{c a-b b} + \frac1{a b-c c}, \tag{1}$$
$$ R_2 := \frac{a}{(b c-a a)^2} + \frac{b}{(c a-b b)^2} + \frac{c}{(a b-c c)^2}, \tag{2}$$
and polynomials
$$ P_1:=a+b+c,\qquad \qquad P_2:=a b+b c+c a,\\ P_3:=P_2\!-\!a a\!-\!b b\!-\!c c,\quad
P_4:=a b c P_1\!-\!a a b b\!-\!b b c c\!-\!c c a a,
\tag{3} $$
and product of polynomials
$$ D := (b c-a a)(c a-b b)(a b-c c). \tag{3} $$
The factorization of $\,R_1,R_2\,$ are
$$ R_1 = \frac{P_2P_3}D, \qquad R_2 = \frac{P_1 P_2 P_3 P_4}{D^2}. \tag{4} $$
The result is that $\,R_1 = 0\,$ if $\,P_2P_3=0\,$ and this also implies
that $\,R_2 = 0.$
