2. If three distinct real numbers $a, b, c$, satisfy $a^{3}+b^{3}+c^{3}=3 a b c$, then $a+b+c=$
$$
\begin{array}{l}
a^{3}+b^{3}+c^{3}-3 a b c=(a+b)^{3}-3 a b(a+b)+c^{3}-3 a b c \\
=(a+b+c)\left[(a+b)^{2}-c(a+b)+c^{2}-3 a b\right]=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-a c-b c\right) \\
=\frac{1}{2}(a+b+c)\left[(a-b)^{2}+(a-c)^{2}+(b-c)^{2}\right]=0 \Rightarrow a+b+c=0 .
\end{array}
$$