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

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