Find $\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$ if $a+b+c=0$ I'm stuck at this algebra problem, it seems to me that's what's provided doesn't even at all.
Provided:  $$a+b+c=0$$
Find the value of:  $$\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$$
Like I'm not sure where to start, and the provided clue doesn't even make sense.  There's no way I can think of to factor this big polynomial into a form like $a\times p+b\times q+c\times r=s$ where $p,q,r,s\in\mathbb{Z}$.
Thanks!

Note that:
$$\frac{a^2}{2a^2+bc}=\frac{a^2}{a^2+a(-b-c)+bc}=\frac{a^2}{(a-b)(a-c)}$$
This applies in the same way for:
$$\frac{b^2}{2b^2+ac}=\frac{b^2}{(b-c)(b-a)}\ \text{and}\ \frac{c^2}{2c^2+ab}=\frac{c^2}{(c-a)(c-b)}$$Therefore, the original equation is equal to:
\begin{align}
&\frac{a^2}{(a-b)(a-c)}+\frac{b^2}{(b-c)(b-a)}+\frac{c^2}{(c-a)(c-b)}
\\
\\
&=\frac{-a^2(b-c)-b^2(c-a)-c^2(a-b)}{(a-b)(b-c)(c-a)}
\\
\\
&=\frac{-a^2b-b^2c-c^2a+a^2c+b^2a+c^2b}{(a-b)(b-c)(c-a)}
\\
\\
&=\frac{(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}
\\
\\
&=\boxed{1}
\end{align}
