3. Let $a, b, c$ satisfy $a+b+c=a^{3}+b^{3}+c^{3}=0, n$ be any real number, then $a^{2 n+1}+b^{2 n+1}+c^{2 n+1}=$
Since $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)=0 \Rightarrow a b c=0$, then $a, b, c$ at least one is 0. Without loss of generality, let $a=0 \Rightarrow c=-b$,
so $a^{2 n+1}+b^{2 n+1}+c^{2 n+1}=0+b^{2 n+1}+(-b)^{2 n+1}=0$.