Question

If $a+b+c$ are all non-zero & $a+b+c=0$ prove that,
$\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}=3$

Answer 1

Given:$a+b+c=0$ where $a\ne 0,b\ne 0,c\ne 0$

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

$\Rightarrow \frac{a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}=3$

$\Rightarrow a^3+b^3+c^3=3abc$

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

Since $a+b+c=0$ we have

$\left(0\right)\times (a^2+b^2+c^2-ab-bc-ca)=3abc$

$\Rightarrow 3abc=0$

$\Rightarrow a^3+b^3+c^3-3abc=0$

or Divide both sides by $abc$ we get

$\Rightarrow \frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}-3=0$

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