Example 2 Given that $\alpha$ is a root of the equation $x^{2}-x-1=0$. Try to find the value of $\alpha^{18}+323 \alpha^{-6}$.

 

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Given: $\because \alpha$ is a root of $x^{2}-x-1=0$,
$$
\therefore \alpha^{2}-\alpha-1=0 \text {. }
$$

Thus, since $\alpha \neq 0$, we have $\alpha-\alpha^{-1}=1$.
Therefore, $\alpha^{2}+\alpha^{-2}=\left(\alpha-\alpha^{-1}\right)^{2}+2=3$,
$$
\begin{array}{l}
\alpha^{4}+\alpha^{-4}=7, \alpha^{6}+\alpha^{-6}=18, \\
\alpha^{12}+\alpha^{-12}=322 .
\end{array}
$$
$$
\begin{aligned}
\text { Hence, the original expression } & =\alpha^{18}+\left(\alpha^{12}+\alpha^{-12}+1\right) \alpha^{-6} \\
& =\alpha^{18}+\alpha^{6}+\alpha^{-18}+\alpha^{-6} \\
& =\left(\alpha^{6}+\alpha^{-6}\right)\left(\alpha^{12}+\alpha^{-12}\right) \\
& =18 \times 322=5796 .
\end{aligned}
$$