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

= $$ (\frac{a^2 + b^2 + 2ab}{ab}) + (\frac{b^2 + c^2 + 2bc}{bc}) + (\frac{c^2 + a^2 + 2ca}{ca}) $$................(1)

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

= $$ (\frac{a}{b} + \frac{b}{a} + 2) + (\frac{b}{c} + \frac{c}{b} + 2) + (\frac{c}{a} + \frac{a}{c} + 2) $$ .......................(2)

= $$ (\frac{b+c}{a}) + (\frac{c+a}{b}) +(\frac{a+b}{c}) + 2 + 2 + 2 $$

using (1) , (2) and (3)

= $$ \frac{-a}{a} + \frac{-b}{b} +  \frac{-c}{c} + 6 $$

= -1-1-1+6 = 6 - 3 = 3