Given $a 、 b 、 c$ are real numbers, and
$$
a^{2}+b^{2}+c^{2}+2 a b=1, a b\left(a^{2}+b^{2}+c^{2}\right)=\frac{1}{8} \text {, }
$$

The roots of the quadratic equation $(a+b) x^{2}-(2 a+c) x-(a+b)=0$ are $\alpha 、 \beta$. Find the value of $2 \alpha^{3}+\beta^{-5}-\beta^{-1}$.
Solution: From the given, we have
$$
\left\{\begin{array}{l}
\left(a^{2}+b^{2}+c^{2}\right)+2 a b=1, \\
2 a b\left(a^{2}+b^{2}+c^{2}\right)=\frac{1}{4} .
\end{array}\right.
$$

Thus, \(a^{2}+b^{2}+c^{2}\) and \(2 a b\) are the roots of the equation \(t^{2}-t+\frac{1}{4}=0\).
Since the roots of the equation \(t^{2}-t+\frac{1}{4}=0\) are \(t_{1}=t_{2}=\frac{1}{2}\), we have
$$
a^{2}+b^{2}+c^{2}=2 a b=\frac{1}{2} \text {. }
$$

Solving this, we get \(a=b= \pm \frac{1}{2}, c=0\).
Thus, the given equation can be transformed into
$$
x^{2}-x-1=0 \text {. }
$$

Since \(\alpha, \beta\) are the roots of equation (1), then \(\alpha+\beta=1\), and
$$
\left\{\begin{array}{l}
\alpha^{2}-\alpha-1=0, \\
\beta^{2}-\beta-1=0 .
\end{array}\right.
$$

From equation (2), we get \(\alpha^{2}=\alpha+1\). Therefore,
$$
\alpha^{3}=\alpha \cdot \alpha^{2}=\alpha(\alpha+1)=\alpha^{2}+\alpha=2 \alpha+1 \text {. }
$$

Clearly, \(\beta \neq 0\).
Dividing both sides of equation (3) by \(\beta\) and \(\beta^{2}\), and rearranging, we get
$$
\frac{1}{\beta}=\beta-1, \frac{1}{\beta^{2}}=1-\frac{1}{\beta}=2-\beta \text {. }
$$
$$
\begin{array}{l}
\text { and } \beta^{-3}=\beta^{-1} \cdot \beta^{-2}=(\beta-1)(2-\beta) \\
=3 \beta-\beta^{2}-2=2 \beta-3, \\
\beta^{-5}=\beta^{-2} \cdot \beta^{-3} \\
=(2-\beta)(2 \beta-3)=7 \beta-2 \beta^{2}-6 \\
=7 \beta-2(\beta+1)-6=5 \beta-8,
\end{array}
$$

Thus, \(2 \alpha^{3}+\beta^{-5}-\beta^{-1}=4(\alpha+\beta)-5=-1\).
(Zhang Zhecai, Shengzhou Middle School, Jiaojiang District, Taizhou City, Zhejiang Province, 318000)