Question

If $a+b+c=0$ then find value of $\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}$

Answer 1

We have

$(a^3+b^3+c^3-3abc)=(a+b+c)\times (a^2+b^2+c^2-ab-bc-ac)$

When $a+b+c=0$ [Given]

then, $a^3+b^3+c^3=3abc$

$\Rightarrow \frac{a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}=3$ [On dividing $abc$ both sides]

$\Rightarrow \frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=3.$