Let ab+bc+ca=0 then prove that 1a2−bc+1b2−ac+1c2−ab=0.
Given (ab+bc+ca)=0 ; ... (1)
Now, from (1) we get,
a2−(bc)=a2+ab+ac
or, (a2−bc)=a(a+b+c) _ _ _ (2)
similarly, b2−ac=b(a+b+c) ... (3)
and c2−ab=c(a+b+c) ...(4)
Now,
1a2−bc+1b2−ac+1c2−ab
=1(a+b+c){1a+1b+1c} [using (4),(2) and (3)]
=(ab+bc+ca)(abc)(a+b+c)=0 [using (1)].