Two quadratic equations with unequal leading coefficients:
$$
(a-1) x^{2}-\left(a^{2}+2\right) x+\left(a^{2}+2 a\right)=0
$$

and $(b-1) x^{2}-\left(b^{2}+2\right) x+\left(b^{2}+2 b\right)=0$
(where $a, b$ are positive integers) have a common root. Find the value of $\frac{a^{b}+b^{a}}{a^{-b}+b^{-a}}$.
Solution 1 From the conditions, we know $a>1, b>1, a \neq b$.
The two roots of (1) are $a, \frac{a+2}{a-1}$, and the two roots of (2) are $b, \frac{b+2}{b-1}$.
$\because a \neq b$,
$\therefore \quad a=\frac{b+2}{b-1}$,
or $b=\frac{a+2}{a-1}$.
From (3) and (4), we get $a b-a-b-2=0$,
which is $(a-1)(b-1)=3$.
Since $a, b$ are both positive integers, we have
$$
\left\{\begin{array}{l}
a-1=1 \\
b-1=3
\end{array} \quad \text { or } \quad \left\{\begin{array}{l}
a-1=3 \\
b-1=1
\end{array}\right.\right.
$$

Solving, we get
$\left\{\begin{array}{l}a=2 \\ b=4\end{array} \quad\right.$ or $\quad\left\{\begin{array}{l}a=4 \\ b=2\end{array}\right.$
Substituting into the expression to be found, we get
$$
\frac{a^{b}+b^{a}}{a^{-b}+b^{-a}}=a^{b} b^{a}=4^{2} \cdot 2^{4}=256 .
$$

Solution 2 From the conditions, we know $a>1, b>1, a \neq b$,
Let the common root of equations (1) and (2) be $x_{0}$, then
$$
\begin{array}{l}
(a-1)\left(x_{0}\right)^{2}-\left(a^{2}+2\right) x_{0}+\left(a^{2}+2 a\right)=0, \\
(b-1)\left(x_{0}\right)^{2}-\left(b^{2}+2\right) x_{0}+\left(b^{2}+2 b\right)=0 .
\end{array}
$$
(3) $\times(b-1)-(4) \times(a-1)$, rearranging gives
$$
(a-b)(a b-a-b-2)\left(x_{0}-1\right)=0 \text {. }
$$

Since $a \neq b$, then $x_{0}=1$, or $a b-a-b-2=0$.
If $x_{0}=1$, substituting into (3) leads to $a=1$, a contradiction. Therefore, $x_{0} \neq 1$.
$\therefore a b-a-b-2=0$, which is $a b=a+b+2$.
If $a>b>1$, then $b=1+\frac{b}{a}+\frac{2}{a}a>1$, we get $a=2, b=4$.
Substituting into the expression to be found, we get
$$
\frac{a^{b}+b^{a}}{a^{-b}+b^{-a}}=\frac{2^{4}+4^{2}}{2^{-4}+4^{-2}}=\frac{16+16}{\frac{1}{16}+\frac{1}{16}}=32 \times 8=256 .
$$