Example 3 Let $a b c \neq 0, a+b+c=a^{2}+b^{2}+c^{2}=2$. Find the value of the algebraic expression $\frac{(1-a)^{2}}{b c}+\frac{(1-b)^{2}}{c a}+\frac{(1-c)^{2}}{a b}$.
Solve: Regarding $a$ and $b$ as the main elements, then
$$
\begin{array}{l}
\left\{\begin{array}{l}
a+b=2-c, \\
a^{2}+b^{2}=2-c^{2} .
\end{array}\right. \\
\text { Hence } a b=\frac{(a+b)^{2}-\left(a^{2}+b^{2}\right)}{2} \\
=c^{2}-2 c+1=(c-1)^{2} \\
\Rightarrow \frac{(1-c)^{2}}{a b}=1 .
\end{array}
$$

Similarly, $\frac{(1-a)^{2}}{b c}=1, \frac{(1-b)^{2}}{c a}=1$.
Thus, $\frac{(1-a)^{2}}{b c}+\frac{(1-b)^{2}}{c a}+\frac{(1-c)^{2}}{a b}=3$.