1. ## Complicated Fractions

Given $a+b+c=0$, evaluate $\frac{1}{b^2+c^2-a^2}+\frac{1}{a^2+c^2-b^2}+\frac{1}{a^2+b^2-c^2}$.

2. Hello,

Let see what $a^2+b^2-c^2$ can give :

$a^2+b^2-c^2=a^2+(b+c)(b-c)$

We know that a+b+c=0 <=> b+c=-a

$a^2+(b+c)(b-c)=a^2-a(b-c)=a(a-b+c)$

As a+c=-b, we finally have :

$a^2+b^2-c^2=-2ab$

Hence we get :

$\frac{1}{a^2+b^2-c^2}=\frac{1}{-2ab}$

With a similar method, we get :

$\frac{1}{a^2+c^2-b^2}=\frac{1}{-2ac}$

$\frac{1}{b^2+c^2-a^2}=\frac{1}{-2bc}$

===> $\frac{1}{b^2+c^2-a^2}+\frac{1}{a^2+c^2-b^2}+\frac{1}{a^2+b^2-c^2} = \frac{1}{-2ab}+\frac{1}{-2ac}+\frac{1}{-2bc}$

$= \frac{c}{-2abc}+\frac{b}{-2abc}+\frac{a}{-2abc} = \frac{a+b+c}{-2abc} = 0$