Question

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

Answer 1

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

Now, from (1) we get,

$a^2-\left(bc\right)=a^2+ab+ac$

or, $(a^2-bc)=a(a+b+c)$ _ _ _ (2)

similarly, $b^2-ac=b(a+b+c)$ ... (3)

and $c^2-ab=c(a+b+c)$ ...(4)

Now,

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

$=\frac{1}{(a+b+c)}\{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\}$ [using (4),(2) and (3)]

$=\frac{(ab+bc+ca)}{\left(abc\right)(a+b+c)}=0$ [using (1)].