$\begin{array}{l}\text { 9. (16 points) Given } a, b, c \in \mathbf{R}, \\ \{a, b, c\}=\left\{-a^{3}, a b c-b^{3}, 2 a b c-c^{3}\right\}. \\ \text { Find }\{a, b, c\}.\end{array}$
Given the conditions:
$$
\left\{\begin{array}{l}
a+b+c=-a^{3}+a b c-b^{3}+2 a b c-c^{3} \\
a b c=\left(-a^{3}\right)\left(a b c-b^{3}\right)\left(2 a b c-c^{3}\right) .
\end{array}\right.
$$

From equation (1), we have:
$$
\begin{array}{l}
\frac{1}{2}(a+b+c)\left((a-b)^{2}+(b-c)^{2}+(c-a)^{2}+2\right) \\
=0 \\
\Rightarrow a+b+c=0 .
\end{array}
$$

Assume \(a b c \neq 0\). From \(a+b+c=0\) and equation (2), we get:
$$
\begin{aligned}
1 & =-a^{2}\left(b^{2}-c a\right)\left(c^{2}-2 a b\right) \\
& =-a^{2}\left((c+a)^{2}-a c\right)\left((a+b)^{2}-2 a b\right) \\
& =-a^{2}\left(a^{2}+a c+c^{2}\right)\left(a^{2}+b^{2}\right) \leqslant 0,
\end{aligned}
$$

which is a contradiction.
Thus, \(a b c=0\).
Therefore, \(\{a, b, c\}=\left\{-a^{3},-b^{3},-c^{3}\right\}\).
Without loss of generality, assume \(c=0\). Then \(b=-a\).
$$
\begin{array}{l}
\text { Hence }\{a,-a\}=\left\{-a^{3}, a^{3}\right\} \Rightarrow a= \pm 1 \\
\Rightarrow\{a, b, c\}=\{-1,0,1\} .
\end{array}
$$

Upon verification, \(\{-1,0,1\}\) satisfies the conditions.