Evaluating $P=\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$ when $a+b+c=0$ 
Let $a,b,c$ such that $$a + b + c =0$$ and 
  $$P=\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$$ is defined.
  Find the value of $P$.

This is a very queer problem.

Put $c=-a-b$ in
$$2a^2+bc=2a^2+b(-a-b)=a^2-ab+a^2-b^2=(a-b)(2a+b)=(a-b)(b-c)$$
Similarly $2b^2+ca=(b-c)(c-a)$ and 
$2c^2+ab=(c-a)(c-b)$
So expression reduce into 
$$\frac{a^2(c-a)+b^2(a-b)+c^2(b-c)}{(a-b)(b-c)(c-a)}=1$$
Factorisation of Numerator
$a^2(c-a)+b^2(a-b)+c^2(b-c)$ is zero when $a=b$ or $b=c$ or $c=a$
$$a^2(b-c)+b^2(c-a)+c^2(a-b)=k(a-b)(b-c)(c-a)$$
For finding $k$ put $a=1,b=2,c=3,$ getting $k=1$
