Question

If $ab+bc+ca=0$ then prove that $\frac{1}{a^2-bc}+\frac{1}{b^2-ca}+\frac{1}{c^2-ab}=0$.

Answer 1

Given,

$ab+bc+ca=0$.....(1).

Now,$\frac{1}{a^2-bc}+\frac{1}{b^2-ca}+\frac{1}{c^2-ab}$

$=\frac{1}{a^2+ab+ac}+\frac{1}{b^2+ab+bc}+\frac{1}{c^2+ac+bc}$ [ Since $-ab=ac+bc,-bc=ab+ac,-ca=ab+bc$ using (1)]

$=\frac{1}{(a+b+c)}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$

$=\frac{1}{(a+b+c)}\left(\frac{ab+bc+ca}{abc}\right)$

$=0$ [Using (1)]