$:$ enderovv B.A.

The numbers $a, b, c$ are such that $a^{2}(b+c)=b^{2}(a+c)=2008$ and $a \neq b$. Find the value of the expression $c^{2}(a+b)$.
$(a-b)(ab+ac+bc)=ab(a-b)+(a^2-b^2)c=a^2(b+c)-b^2(a+c)=0$. Since $a \neq b$, then $ab+ac+bc=0$.

Multiplying by $a-c$, we have

$(a-c)(ac+ab+bc)=a^2(b+c)-c^2(a+b)=0$. Therefore, $c^2(a+b)=a^2(b+c)=2008$.

## Answer

2008.