Let $a, b, c, d$ be the four roots of $X^{4}-X^{3}-X^{2}-1$. Calculate $P(a)+P(b)+P(c)+$ $P(d)$, where $P(X)=X^{6}-X^{5}-X^{4}-X^{3}-X$.
First, note that $P(X)=X^{2}\left(X^{4}-X^{3}-X^{2}-1\right)-X^{3}+X^{2}-X$. Using the recurrence formula for symmetric polynomials, it suffices to compute $-S_{3}+S_{2}-S_{1}$. Now, $\sigma_{1}=1, \sigma_{2}=-1$ and $\sigma_{3}=0$. Therefore, $S_{1}=\sigma_{1}=1, S_{2}=\sigma_{1} S_{1}-2 \sigma_{2}=3$ and $\mathrm{S}_{3}=\sigma_{1} \mathrm{~S}_{2}-\sigma_{2} \mathrm{~S}_{1}+3 \sigma_{3}=4$, and the desired sum is -2.