Finding the sum of the squares of the roots of a quintic

Question

I would like to solve the following problem.

$a, b, c, d, e$ are the roots of $P(x) = x^5 − 160x − 128 = 0$. Compute $a^5+b^5+c^5+d^5+e^5$.

The first thing that came to mind was expand $(a+b+c+d+e)^5$ and subtract all terms from the expression until I got $a^5+b^5+c^5+d^5+e^5$. This method works for finding $a^2+b^2$ for a quadratic. However, I don't want expand $(a+b+c+d+e)^5$.

Is there a way to avoid that?

Answer 1

Since $a$ is rot of $P(x) = x^5 − 160x − 128$ we have $P(a)=0$ so $ a^5 = 160x + 128$ and the same for the other roots. So we have:

$$a^5+...+e^5 = 160 (a+...+e)+5\cdot 128 = 0 +640 =640$$

$a+...+e = 0$ by Vieta's formula.