$\{a,b,c\}\subset \Bbb R$, $a\not =b$ and $a^2(b+c)=b^2(a+c)=2010$. Find $c^2(a+b)$ 
Consider that $\{a,b,c\}\subset \Bbb R$, with $a\not =b$, and it is known that $a^2(b+c)=b^2(a+c)=2010$.
Find $c^2(a+b)$.

A question from the third phase of OBM 2010 (Brazilian Math Olympiad). Sorry if it is a duplicate. My developments are leading to some complicated expressions... probably a wrong approach. Hints and answers are welcomed.

We have 
\begin{eqnarray*} 
a^2b +a^2 c =2010 \\
a b^2+b^2 c =2010.
\end{eqnarray*}
Subtract these equations ( and use $ a \neq b$) we have $ab+bc+ca=0$ so 
\begin{eqnarray*} 
c= - \frac{ab}{a+b}. \\
\end{eqnarray*}
Now multiply the first equation by $b^2$ and the second by $a^2$ and subtract ( and again use $ a \neq b$), we have
\begin{eqnarray*} 
a^2b^2 =2010 (a+b). \\
\end{eqnarray*}
Now 
\begin{eqnarray*} 
c^2 (a+b)=  \frac{a^2b^2}{a+b}= \color{blue}{2010}. \\
\end{eqnarray*}
