$ \frac1{bc-a^2} + \frac1{ca-b^2}+\frac1{ab-c^2}=0$ implies that $ \frac a{(bc-a^2)^2} + \frac b{(ca-b^2)^2}+\frac c{(ab-c^2)^2}=0$

Question

$a ,b , c$ are real numbers such that $ \dfrac1{bc-a^2} + \dfrac1{ca-b^2}+\dfrac1{ab-c^2}=0$ , then how do we prove

(without routine laborious manipulation) that $\dfrac a{(bc-a^2)^2} + \dfrac b{(ca-b^2)^2}+\dfrac c{(ab-c^2)^2}=0$

Answer 1

Multiply out denominators, and rearrange the first equation to $(ab+bc+ca)^2=(ab+bc+ca)(a^2+b^2+c^2)$

Answer 2

$$\frac 1 {b c - a^2 } + \frac 1 {c a - b^2} + \frac 1 {a b - c^2} = \frac { (b c + a c + a b) (bc+ac+ab-a^2-b^2-c^2) } { (bc-a^2)(ac-b^2)(ab-c^2) } $$

$$ \frac a {(bc - a^2)^2} + \frac b {(ca - b^2)^2} + \frac c {(ab - c^2)^2} = \frac{ (bc+ac+ab)(bc+ac+ab-a^2-b^2-c^2) } { (bc-a^2)(ac-b^2)(ab-c^2) } \times \frac{ (a+b+c)(a^2bc+ab^2c+abc^2-b^2c^2-a^2c^2-a^2b^2) }{ (bc-a^2)(ac-b^2)(ab-c^2) } $$