= Root power-sum recurrence
{title2=$p_m-e_1p_{m-1}+\cdots+(-1)^ne_np_{m-n}=0\quad(m>n)$}
For the <monic polynomial> $q(z)=z^n-e_1z^{n-1}+\cdots+(-1)^ne_n$, its root power sums satisfy
$$
p_m=e_1p_{m-1}-e_2p_{m-2}+\cdots+(-1)^{n+1}e_np_{m-n},\qquad m>n.
$$
Each root obeys $q(x_i)=0$. Multiplying that equation by $x_i^{m-n}$ and summing over root positions proves the <recurrence relation>. The first $n$ power sums are initialized by <Newton's identities>; their later values require only this fixed-order recurrence. For $q(z)=z^3-6z^2+11z-6$, it gives $p_4=6\cdot36-11\cdot14+6\cdot6=98$. Directly, $1^4+2^4+3^4=98$. No new root-solving step is needed.<Rational Univariate Representation (Lecture 12) by Jan Verschelde>[§4, p. 3, equations (21)–(33)]{ref}
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