Example 12 If $\alpha, \beta, \gamma$ are the roots of the equation $x^{3}-x-1=0$, find the value of $\frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\frac{1+\gamma}{1-\gamma}$.
Given that $x^{3}-x-1=(x-\alpha)(x-\beta)(x-\gamma)$, then $\alpha+\beta+\gamma=0, \beta \gamma+\gamma \alpha+\alpha \beta=$ $-1, \alpha \beta \gamma=1$.
$$
\begin{array}{l}
\therefore \frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\frac{1+\gamma}{1-\gamma} \\
=2\left(\frac{1}{1-\alpha}+\frac{1}{1-\beta}+\frac{1}{1-\gamma}\right)-3 \\
=2 \times \frac{3-2(\alpha+\beta+\gamma)+(\beta \gamma+\gamma \alpha+\alpha \beta)}{(1-\alpha)(1-\beta)(1-\gamma)}-3 \\
=2 \times \frac{3-2 \times 0-1}{1-0-1-1}-3=-?
\end{array}
$$