Find the limit of some roots raised to n let $x_1,x_2,x_3$ be the roots of $x^3-x^2-1=0$. If $x_1$ is a real root of the equation, compute: $\lim_{n\to\infty}(x_2^n+x_3^n).$ First I find these relations using viete: $x_1+x_2+x_3=1$ $x_1^2+x_2^2+x_3^2=1$ $x_1^3+x_2^3+x_3^3=4$ Now, we can find a recurrence of the other sum of roots using the inial equation and we get this sequence: $$a_n=a_{n-1}+a_{n-3}$$ where $$a_1=1, a_2=1, a_3=4$$ And I tried something like this to try and solve this problem but couldn't work it out until the end... I'm actually stuck here.

Standard calculus proves that $x^3-x^2-1$ has a single real root $x_1$, and $x_1>1$. It is also standard that the polynomial has two other roots $x_2$ and $x_3$ which are complex conjugates: $x_2=\overline{x_3}$, thus $|x_2|=|x_3|$.

By Vieta, $x_1x_2x_3=1$, thus $|x_2|^2=|x_2x_3|=\frac{1}{|x_1|}<1$, hence $|x_2|<1$.

Let $x_2 = re^{i\theta}$ and note that $x_2^n+x_3^n = r^n 2\cos(n\theta)$

But $r=|x_2|<1$, thus $\lim_n r^n 2\cos(n\theta) = 0$.

Hence $\lim_n x_2^n+x_3^n = 0$.