Algebraic Manipulations within a Polynomial

Question

15. Let $P(x) = x^3 − ax^2 + x − 1$, for a positive integer $a$. If the sum of the cubes of the roots of $P$ is $113$, what is the sum of the fourth powers of the roots of $P$?

I know the answer is $537$, but I don't know how to solve the question. I don't even know where to begin...

Answer 1

The sums of powers of roots $s_n$ follow the recursion $$ s_{n+3}-as_{n+2}+s_{n+1}-s_n=0. $$ Trivially $s_0=3$. By the Viete identities, $s_1=a$ and $s_2=a^2-2$. Then as given $$ 113=s_3=as_2-s_1+s_0=a^3-3a+3 $$ with the only real root $a=5$. The number to be found is $$ s_4=as_3-s_2+s_1=5\cdot113-23+5=547 $$