1. Given $a^{2}(b+c)=b^{2}(a+c)=2010$, and $a \neq b$. Then $c^{2}(a+b)=$
1. 2010 .

Notice
$$
\begin{array}{l}
a^{2}(b+c)-b^{2}(a+c) \\
=(a-b)(a c+b c+a b)=0 .
\end{array}
$$

Since $a \neq b$, it follows that $a c+b c+a b=0$.
$$
\begin{array}{l}
\text { Then } c^{2}(a+b)-b^{2}(a+c) \\
=(c-b)(a c+b c+a b)=0 .
\end{array}
$$

Therefore, $c^{2}(a+b)=b^{2}(a+c)=2010$.