Roots--4th power

let $f(x) = x^3-2x^2+x-1$ ,

let $a, b, c$ be the roots of $f(x)=0$ , and define $s_n = a^n+b^n+c^n$

because $f(a)=f(b)=f(c)=0$ , we know that $a^nf(a)+b^nf(b)+c^nf(c)=0$

rearranging the terms we get $a^nf(a)+b^nf(b)+c^nf(c) = s_{n+3}-2s_{n+2}+s_{n+1}-s_n$

and thus $s_{n+3} = 2s_{n+2}-s_{n+1}+s_{n}$

using this identity

$s_4 = 2s_{3}-s_{2}+s_{1} = 2(2s_{2}-s_{1}+s_{0})-s_{2}+s_{1} = 3s_2 - s_1 + 2s_0$

substituding $s_0=3$ , $s_1=2$ , and $s_2 = (a+b+c)^2-2(ab+bc+ac)=2^2-2=2$

we finally have $s_4 = 3\times 2 - 2 + 2\times 3 = \boxed{10}$ , which is what we wanted.