3. Given $\frac{a^{3}+b^{3}+c^{3}-3 a b c}{a+b+c}=3$. Then $(a-b)^{2}+(b$ $-c)^{2}+(a-b)(b-c)$ is $(\quad)$.
(A) 1
(B) 2
(C) 3
(D) 4
3. C.
$$
\begin{aligned}
3 & =\frac{a^{3}+b^{3}+c^{3}-3 a b c}{a+b+c} \\
& =\frac{(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)}{a+b+c} \\
& =(a-b)^{2}+(b-c)^{2}+(a-b)(b-c) .
\end{aligned}
$$