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.
 A: Elegant approach 
Think of $p(a, b, c)$ as a function $p : X\to\mathbb{R}$, where $X = \{(a, b, c)\in\mathbb{R}^3 : a+b+c=0\}$.
If $p$ is a constant function, i.e. if $p(a, b, c)\equiv P$ for some constant $P$, then in particular, $p(a, b, c) = P$ for any values of $a, b, c$ we might care to choose, so long as they satisfy $a+b+c=0$.
So it's as easy as just picking some set of values that will be easy to evaluate, such as $a=1, b=-1, c=0$, and substituting:
\begin{align}
  p(a, b, c) &= \frac{a^2}{2a^2+bc} + \frac{b^2}{2b^2+ac} + \frac{c^2}{2c^2+ab} \\ \Rightarrow
  p(1, 0, 0) &= \frac{1}{2+0} + \frac{(-1)^2}{2(-1)^2+0} + 0 = \frac{1}{2} + \frac{1}{2} = 1
\end{align}
Therefore $P = 1$.
(In case you don't want to accept from the statement of the problem that $p$ is indeed a constant function, it's straightforward enough to see: because $c = -a-b$, any increment in $a$ will change $p$ by $dp = (\partial{p}/\partial{a} - \partial{p}/\partial{c})\,da$; on the other hand, the symmetry of $p(a, b, c)$ in all three arguments implies $p(a, b, c) = p(c, b, a) \Rightarrow$ $\partial{p}/\partial{a} - \partial{p}/\partial{c} = 0 \Rightarrow$ $dp/da=0$. A similar argument can be made for $b$, to establish $dp/db=0$.)

Brute force approach 
If you prefer to ignore the easy route shown above, and brute force this problem, then that is also possible. From $a + b + c = 0$, substitute $c = -(a + b)$, and simplify:
\begin{align}
  P &= \frac{a^2}{2a^2+bc} + \frac{b^2}{2b^2+ac} + \frac{c^2}{2c^2+ab}  \\
    &= \frac{a^2}{2a^2-ab-b^2} + \frac{b^2}{2b^2-ab-a^2} + \frac{(a+b)^2}{(2a^2+4ab+2b^2)+ab}  \\
    &= \frac{a^2}{(a-b)(2a+b)} + \frac{-b^2}{(a-b)(a+2b)} + \frac{(a+b)^2}{(2a+b)(a+2b)}  \\
    &= \frac{a^2(a+2b) - b^2(2a+b) + (a+b)(a^2-b^2)}{(a-b)(2a+b)(a+2b)}  \\
    &= \frac{(a^3+2a^2b) + (-2ab^2-b^3) + (a^3-ab^2 + a^2b-b^3)}{(a-b)(2a+b)(a+2b)}  \\
    &= \frac{2a^3+3a^2b -3ab^2-2b^3}{(a-b)(2a+b)(a+2b)}  
     = \frac{(a-b)(2a+b)(a+2b)}{(a-b)(2a+b)(a+2b)} = 1
\end{align}
A: 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 $k=1$
