1. (1996 28th Canadian Mathematical Olympiad) If $\alpha, \beta, \gamma$ are the roots of the equation $x^{3}-x-1=0$, find the value of $\frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\frac{1+\gamma}{1-\gamma}$.
1. From the given, we have
$$
x^{3}-x-1=(x-\alpha)(x-\beta)(x-\gamma) \text {, }
$$

Therefore,
$$
\left\{\begin{array}{l}
\alpha+\beta+\gamma=0, \\
\alpha \beta+\beta \gamma+\gamma \alpha=-1, \\
\alpha \beta \gamma=1 .
\end{array}\right.
$$

Thus, we have
$$
\begin{aligned}
& \frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\frac{1+\gamma}{1-\gamma}=2\left(\frac{1}{1-\alpha}+\frac{1}{1-\beta}+\frac{1}{1-\gamma}\right)-3 \\
= & 2\left(\frac{3-2(\alpha+\beta+\gamma)+(\alpha \beta+\beta \gamma+\gamma \alpha)}{(1-\alpha)(1-\beta)(1-\gamma)}\right)-3 \\
= & 2\left(\frac{3-1}{1-1-1}\right)-3=-7 .
\end{aligned}
$$