$\{a,b,c\}\subset \Bbb R$, $a\not =b$ 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.
 A: 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*}
A: Observe that:
$$
a^2(b+c)-b^2(a+c)=0
$$
Expanding it, we get
$$
a^2b+a^2c-b^2a-b^2c=0
$$
Factoring we get,
\begin{align}
ab(a-b)+c(a^2-b^2)&=0\\
ab(a-b)+c(a-b)(a+b)&=0\\
(a-b)(ab+c(a+b))&=0.
\end{align}
Since $a\not=b$, it follows that 
$$
c(a+b)=-ab
$$
or that
$$
c^2(a+b)=-abc.
$$
Take the first equation
$$
a^2(b+c)=2010
$$
and expand it:
$$
a^2b+a^2c=2010.
$$
Substitute $-c(a+b)=ab$ to get
$$
-ac(a+b)+a^2c=2010.
$$
Expanding this, we get
$$
-a^2c-abc+a^2c=2010.
$$
Therefore,
$$
-abc=2010,
$$
which we know is
$$
c^2(a+b).
$$
A: we have $$c=\frac{2010}{a^2}-b$$ and $$c=\frac{2010}{b^2}-a$$ from here we get
$$2010(b^2-a^2)=a^2b^2(b-a)$$ or $$b+a=\frac{a^2b^2}{2010}$$ and since $$c^2(a+b)=-abc$$ we get
$$b+a=\frac{\frac{2010^2}{c^2}}{2010}$$ therefore $$c^2(a+b)=2010$$
