1. Given $a b c \neq 0$, and
$$
a+b+c=a^{2}+b^{2}+c^{2}=2 \text {. }
$$

Then the value of the algebraic expression $\frac{(1-a)^{2}}{b c}+\frac{(1-b)^{2}}{c a}+\frac{(1-c)^{2}}{a b}$ is ( ).
(A) 3
(B) -3
(C) 1
(D) -1
- 1.A.

Since $a+b+c=a^{2}+b^{2}+c^{2}=2$, we have
$$
\begin{array}{l}
a b+b c+c a \\
=\frac{1}{2}\left[(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)\right]=1
\end{array}
$$

Notice that
$$
\begin{array}{l}
a(1-a)^{2}=a\left(1-2 a+a^{2}\right) \\
=a\left[(a b+b c+c a)-(a+b+c) a+a^{2}\right] \\
=a b c .
\end{array}
$$

Similarly, $b(1-b)^{2}=a b c, c(1-c)^{2}=a b c$.
$$
\begin{array}{l}
\text { Hence } \frac{(1-a)^{2}}{b c}+\frac{(1-b)^{2}}{c a}+\frac{(1-c)^{2}}{a b} \\
=\frac{a(1-a)^{2}+b(1-b)^{2}+c(1-c)^{2}}{a b c} \\
=\frac{3 a b c}{a b c}=3 .
\end{array}
$$