1. The numbers $u, v, w$ are roots of the equation $x^{3}-3 x-1=0$. Find $u^{9}+v^{9}+w^{9}$. (12 points)
Solution. $\quad$ According to Vieta's theorem, $u+v+w=0, uv+vw+uw=-3, uvw=1$. Consider the sequence $S_{n}=u^{n}+v^{n}+w^{n}$. We have $S_{0}=3, S_{1}=0$. Let's find $S_{2}$: $S_{2}=u^{2}+v^{2}+w^{2}=(u+v+w)^{2}-2(uv+vw+uw)=6. \quad$ Since $\quad u^{3}=3u+1, v^{3}=$ $3v+1, w^{3}=3w+1, \quad$ then $S_{3}=u^{3}+v^{3}+w^{3}=3S_{1}+S_{0}$. Further, $u^{n}=3u^{n-2}+u^{n-3}, v^{n}=$ $3v^{n-2}+v^{n-3}, w^{n}=3w^{n-2}+w^{n-3}$, then $S_{n}=3S_{n-2}+S_{n-3}$. We find $S_{3}=3, S_{4}=18, S_{5}=$ $15, S_{6}=57, S_{7}=63, S_{9}=246$

## Answer: 246