Example 1 Constructing equations to solve problems:
Let $x=\frac{\sqrt{5}+1}{2}$, then $\frac{x^{3}+x+1}{x^{5}}=$ $\qquad$ $(1988$, Shanghai Junior High School Mathematics Competition)
If we substitute $x=\frac{\sqrt{5}+1}{2}$ directly, it is a clumsy method, leading to complicated calculations. If we regard $\frac{\sqrt{5}+1}{2}$ as a root of a quadratic equation, then the other root is $\frac{1-\sqrt{5}}{2}$. Thus, we can construct the equation
$$
x^{2}-x-1=0 \text {. }
$$

Let one root of this equation be $\alpha=\frac{\sqrt{5}+1}{2}$, and the other root be $\beta=\frac{1-\sqrt{5}}{2}$. Then
$$
\begin{array}{l}
\alpha^{2}-\alpha-1=0, \\
\alpha^{2}=\alpha+1, \\
\alpha^{3}=\alpha(\alpha+1)=\alpha^{2}+\alpha, \\
\alpha^{3}+\alpha+1=\alpha^{2}+2 \alpha+1=(\alpha+1)^{2}=\alpha^{4} . \\
\therefore \quad \frac{x^{3}+x+1}{x^{5}}=\frac{\alpha^{4}}{\alpha^{5}}=\frac{1}{\alpha} . \\
\because \alpha \beta=-1, \\
\therefore \frac{1}{\alpha}=-\beta=\frac{1}{2}(\sqrt{5}-1) .
\end{array}
$$

For equation (1), we can derive the following problems, which require the use of Vieta's formulas.