Question

If $a+b+c=0$ , find value of
$\frac{(b+c{)}^2}{bc}+\frac{(c+a{)}^2}{ca}+\frac{(a+b{)}^2}{ab}$

Answer 1

Given, $a+b+c=0$

$\Rightarrow b+c=-a$ and $c+a=-b$ and $a+b=-c$

Let $x=\frac{(b+c{)}^2}{bc}+\frac{(c+a{)}^2}{ca}+\frac{(a+b{)}^2}{ab}$

$=\frac{(-a{)}^2}{bc}+\frac{(-b{)}^2}{ca}+\frac{(-c{)}^2}{ab}$

$=\frac{(a{)}^2}{bc}+\frac{(b{)}^2}{ca}+\frac{(c{)}^2}{ab}$

$=\left(\frac{a^3+b^3+c^3}{abc}\right)$

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

$\therefore x=3$