Example 7 Two quadratic equations with unequal leading coefficients $(a-1) x^{2}-\left(a^{2}+2\right) x+\left(a^{2}+2 a\right)=0$, (1) and $(b-1) x^{2}-\left(b^{2}+2\right) x+\left(b^{2}+2 b\right)=0$ ( $a, b$ are positive integers) have a common root. Find the value of $\frac{a^{b}+b^{a}}{a^{-b}+b^{-a}}$.
Given the conditions $a>1, b>1, a \neq b$.
Assume the common root of equations (1) and (2) is $x_{0}$. Then
$$
\begin{array}{l}
(a-1) x_{0}^{2}-\left(a^{2}+2\right) x_{0}+\left(a^{2}+2 a\right)=0, \\
(b-1) x_{0}^{2}-\left(b^{2}+2\right) x_{0}+\left(b^{2}+2 b\right)=0 .
\end{array}
$$
(3) $\times(b-1)$ - (4) $\times(a-1)$ and simplifying, we get
$$
(a-b)(a b-a-b-2)\left(x_{0}-1\right)=0 \text {. }
$$

Since $a \neq b$, we have
$x_{0}=1$ or $a b-a-b-2=0$.
If $x_{0}=1$, substituting into equation (3) yields $a=1$, which is a contradiction.
Therefore, $x_{0} \neq 1$.
Hence, $a b-a-b-2=0$
$\Rightarrow(a-1)(b-1)=3$
$\Rightarrow a=2, b=4$ or $a=4, b=2$.
Substituting into the required algebraic expression, we get
$$
\frac{a^{b}+b^{a}}{a^{-b}+b^{-a}}=\frac{2^{4}+4^{2}}{2^{-4}+4^{-2}}=256 .
$$