Sum of cubes of roots of a quartic equation

This is precisely what Newton's Sums are for.

Let $S_n$ be the sum of the $n$-th powers of the roots of $a_nx^n+a_{n-1}x^{n-1} + \cdots + a_1x+a_0$.

Then, Newton's Sums tells us that $\displaystyle\sum_{k = 0}^{m-1}\left(a_{n-j}S_{m-j}\right)+ma_{n-m} = 0$ for each positive integer $m$.

We can write out the first few equations explicitly as follows:

$a_nS_1 + a_{n-1} = 0$

$a_nS_2 + a_{n-1}S_1 + 2a_{n-2} = 0$

$a_nS_3 + a_{n-1}S_2 + a_{n-2}S_1 + 3a_{n-3} = 0$

$a_nS_4 + a_{n-1}S_3 + a_{n-2}S_2 + a_{n-3}S_1 + 4a_{n-4} = 0$

and so on. Note that we define $a_j = 0$ for $j < 0$.

Applying those formula to the polynomial $1x^4+0x^3-5x^2+2x+1$ yields the following:

$1 \cdot S_1 + 0 = 0 \leadsto S_1 = 0$ (sum of roots)

$1 \cdot S_2 + 0 \cdot S_1 + 2 \cdot -5 = 0 \leadsto S_2 = 10$ (sum of squares of roots)

$1 \cdot S_3 + 0 \cdot S_2 + - 5 \cdot S_1 + 3 \cdot 2 = 0 \leadsto S_3 = -6$ (sum of cubes of roots)

and so on.