6. Given $a+b+c=3, a^{2}+b^{2}+c^{2}=3$. Then the value of $a^{2003}+b^{2003}+c^{2003}$ is ( ).
(A) 0
(B) 3
(C) 6
(D) $3^{2003}$
6.B.
$$
\begin{array}{l}
(a-1)^{2}+(b-1)^{2}+(c-1)^{2} \\
=\left(a^{2}+b^{2}+c^{2}\right)-2(a+b+c)+3=0,
\end{array}
$$

Then $a=1, b=1, c=1$. Therefore, $a^{2 \alpha B}+b^{2 \alpha B}+c^{2 \alpha B}=3$.