19. Let $a, b, c, d, e$ be five numbers satisfying the following conditions:
$$
\begin{array}{l}
a+b+c+d+e=0, \quad \text { and } \\
a b c+a b d+a b e+a c d+a c e+a d e+b c d+b c e+b d e+c d e=2008 .
\end{array}
$$

Find the value of $a^{3}+b^{3}+c^{3}+d^{4}+e^{3}$.
19. Answer: 6024 .
Note that $(a+b+c+d+e)^{3}=a^{3}+b^{3}+c^{3}+d^{3}+e^{3}+3 a^{2}(b+c+d+e)+3 b^{2}(c+d+e+$ a) $+3 e^{2}(d+e+a+b)+3 d^{2}(e+a+b+c)+3 e^{2}(a+b+c+d)+6(a b c+a b d+a b e+\cdots)$.
Since $a+b+c+d+e=0 . b+c+d+e=-a$, so $3 a^{2}(b+c+d+e)=-3 a^{3}$. Similarly, we get $-3 b^{3},-3 c^{3},-3 d^{3}$ and $-3 e^{3}$. Thus, we have $2\left(a^{3}+b^{3}+c^{3}+d^{4}+e^{3}\right)=$ $6(a b c+a b d+a b c+\cdots)$, so $a^{3}+b^{3}+c^{3}+d^{3}+e^{3}=3 \times 2008=6024$.