1. Given $\frac{a^{3}+b^{3}+c^{3}-3 a b c}{a+b+c}=3$. Then the value of $(a-b)^{2}+(b-c)^{2}+(a-b) \cdot(b-c)$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
,$- 1 .(\mathrm{C})$.
$$
\begin{array}{l}
\because a^{3}+b^{3}+c^{3}-3 a b c \\
=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) .
\end{array}
$$
$\therefore$ The original condition equation is $a^{2}+b^{2}+c^{2}-a b-b c-c a=3$. The required algebraic expression, when expanded and combined, is
$$
a^{2}+b^{2}+c^{2}-a b-b c-c a=3 .
$$