If $a+b+c=0$, then what is the value of $\frac{(b+c)^2}{b c}+\frac{(c+a)^2}{c a}+\frac{(a+b)^2}{a b}$ ?

1 -3 -1 3

3

$\frac{(b+c)^2}{b c}+\frac{(c+a)^2}{c a}+\frac{(a+b)^2}{a b}$ = Let the value of a = 2, b = -1 and c = -1 and put them in required equation, = $\frac{(-1-1)^2}{-1 \times 1}+\frac{(-1+2)^2}{-1 \times 2}+\frac{(2 - 1)^2}{2 \times -1}$ = 4 - \(\frac{1}{2}\) - \(\frac{1}{2}\) = 3