Example 6. Two quadratic equations with unequal leading coefficients
$$
\begin{array}{l}
(a-1) x^{2}-\left(a^{2}+2\right) x+\left(a^{2}+2 a\right)=0, \quad (1) \\
(b-1) x^{2}-\left(b^{2}+2\right) x+\left(b^{2}+2 x\right)=0
\end{array}
$$
(Where $a, b$ are positive integers) have a common root. Find the value of $\frac{a^{b}+b^{2 a}}{a^{-b}+b^{-a}}$. (1986, National Junior High School Competition)
Analyzing, since the two equations have a common root (known number), let's denote it as $x_{0}$. Clearly, $x_{0} \neq 1$ (otherwise $a=b$), so the equation in terms of the unknown $x$ can be transformed into equations in terms of $a$ and $b$:
$$
\left(1-x_{0}\right) a^{2}+\left(x_{0}^{2}+2\right) a-\left(x_{0}^{2}+2 x_{0}\right)=0,
$$
$$
\left(1-x_{0}\right) b^{2}+\left(x_{0}^{2}+2\right) b-\left(x_{0}^{2}+2 x_{0}\right)=0 \text {. }
$$

It is evident that $a, b$ are the two distinct positive integer roots of the equation
$$
\left(1-x_{0}\right) y^{2}+\left(x_{0}^{2}+2\right) y-\left(x_{0}^{2}+2 x_{0}\right)=0
$$

By Vieta's formulas, we get
$$
a+b=\frac{x_{0}^{2}+2}{x_{0}-1}, a b=2+\frac{x_{0}^{2}+2}{x_{0}-1} .
$$

Thus, $a b=2+(a+b)$.
If $a>b>1$, then $b=1+\frac{2}{a}+\frac{b}{a}a>1$, leading to $a=2, b=4$. Therefore,
$$
\frac{a^{b}+b^{a}}{a^{-b}+b^{-a}}=a^{b} b^{a}=2^{4} \times 4^{2}=256 .
$$