Example 2 If $a+b+c=a b c \neq 0$, find the value of $\frac{\left(1-a^{2}\right)\left(1-b^{2}\right)}{a b}+\frac{\left(1-b^{2}\right)\left(1-c^{2}\right)}{b c}+$ $\frac{\left(1-c^{2}\right)\left(1-a^{2}\right)}{a c}$.
(1990, Wuhan City Mathematics Competition)
$\begin{array}{l}\text { Solution: Original expression }=\frac{1-a^{2}-b^{2}+a^{2} b^{2}}{a b} \\ +\frac{1-b^{2}-c^{2}+b^{2} c^{2}}{b c}+\frac{1-a^{2}-c^{2}+a^{2} c^{2}}{a c} \\ =\frac{1}{a b}-\frac{a}{b}-\frac{b}{a}+a b+\frac{1}{b c}-\frac{b}{c}-\frac{c}{b}+b c \\ +\frac{1}{a c}-\frac{a}{c}-\frac{c}{a}+a c \\ =\left(\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}\right)-\frac{b+c}{a}-\frac{a+c}{b}-\frac{a+b}{c} \\ +a b+b c+c a \\ =\frac{a+b+c}{a b c}-(b c-1)-(a c-1) \\ -(a b-1)+a b+b c+c a \\ =4 \text {. } \\\end{array}$