Prove that $\alpha^n+\beta^n+\gamma^n \equiv 4^n+5^n+(-6)^n \pmod{17}$ 
Let $\alpha,\beta,\gamma$ be the roots of $x^3-3x^2+1 = 0$. Prove that $$\alpha^n+\beta^n+\gamma^n \equiv 4^n+5^n+(-6)^n \pmod{17}$$ where $n$ is an integer.

It sort of makes sense why they are congruent since we can say $x^3-3x^2+1 \equiv (x-4)(x-5)(x+6) \pmod{17}$, but since $\alpha,\beta,\gamma$ aren't integers, how do we prove this?

Newton's identities for a cubic polynomial suggests that in this scenario $P_n = \alpha^n + \beta^n + \gamma^n = 3P_{n-1} - P_{n-3}$, and $P_0 = 3, P_1 = 3, P_2 = 9$.
For the other root sum it suggests polynomial $(x-4)(x-5)(x+6) = x^3-3 x^2-34 x+120$ which has recurrence $P_n = 4^n + 5^n + (-6)^n = 3P_{n-1} - 34P_{i-2} - 120P_{n-3}$, and $P_0 = 3, P_1 = 3, P_2 = 77$.
Taken mod $17$ you get $3P_{n-1} - P_{n-3}$ with $P_0 = 3, P_1 = 3, P_2 = 9$, the same recurrence as before.
