Let $P(x)=x^{3}-3 x-1$. Denote $x_{1}, x_{2}$ and $x_{3}$ as its three real roots. Evaluate the quantity $x_{1}^{3}+x_{2}^{3}+x_{3}^{3}$.
A first solution is to use the Gauss identity

$$
x_{1}^{3}+x_{2}^{3}+x_{3}^{3}-3 x_{1} x_{2} x_{3}=\left(x_{1}+x_{2}+x_{3}\right)\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-x_{1} x_{2}-x_{2} x_{1}-x_{1} x_{3}\right)
$$

which leads to

$$
x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=3 \sigma_{3}+\sigma_{1}\left(\sigma_{1}^{2}-3 \sigma_{2}\right)=3
$$

since $\sigma_{1}=0$ and $\sigma_{3}=1$ according to the relations between coefficients and roots.

Alternatively, a more ingenious solution is to exploit the fact that $P\left(x_{1}\right)=P\left(x_{2}\right)=P\left(x_{3}\right)=0$, so for $i \in\{1,2,3\}$, we have $x_{i}^{3}=3 x_{i}+1$. We deduce that $x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=3\left(x_{1}+x_{2}+\right.$ $\left.x_{3}\right)+3=3$.