Let $\alpha, \beta, \gamma$ be the three roots of $x^{3}-x-1$, calculate: $\frac{1-\alpha}{1+\alpha}+\frac{1-\beta}{1+\beta}+\frac{1-\gamma}{1+\gamma}$
A safe method, even if in this case we can find a faster one, is to look for the equation whose roots are $\frac{1-\alpha}{1+\alpha}, \frac{1-\beta}{1+\beta}, \frac{1-\gamma}{1+\gamma}$ and to calculate the sum of the roots of this equation from its coefficients. If $x$ is a root of $x^{3}-x-1$, what equation is $y=\frac{1-x}{1+x}$ a root of? We note that $x=\frac{1-y}{1+y}$ (the function is involutive), so $\left(\frac{1-y}{1+y}\right)^{3}-\left(\frac{1-y}{1+y}\right)-1=$ 0, which means: $(1-y)^{3}-(1-y)(1+y)^{2}-(1+y)^{3}=0$. The equation in $y$ is therefore: $-y^{3}+y^{2}-7 y-1$, the sum of its roots is 1.