Root power-sum recurrence ($p_m-e_1{p}_{m-1}+\cdots +(-1{)}^n{e}_n{p}_{m-n}=0(m>n)$)

For the monic polynomial $q(z)=z^n-e_1{z}^{n-1}+\cdots +(-1{)}^n{e}_n$, its root power sums satisfyEach 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 $p_4=6\cdot 36-11\cdot 14+6\cdot 6=98$. Directly, $1^4+2^4+3^4=98$. No new root-solving step is needed.^[ref]