Newton's identities for three variables

Root equation. For $i=1,2,3$,
$$x_i^3+\alpha x_i^2+\beta x_i+\gamma=0.$$

Sum over $i$. Adding the three equations,
$$\sum_i x_i^3+\alpha\sum_i x_i^2+\beta\sum_i x_i+3\gamma=0,$$
i.e.
$$p_3+\alpha p_2+\beta p_1+3\gamma=0.$$

Convert to $e$-notation. Use $\alpha=-e_1$, $\beta=e_2$, $\gamma=-e_3$:
$$p_3-e_1p_2+e_2p_1-3e_3=0.$$

Solve for $p_3$.
$$p_3=e_1p_2-e_2p_1+3e_3.$$

Check. With $p_1=e_1$ and $p_2=e_1^2-2e_2$ (the $k=1$ and $k=2$ cases of the same recurrence), the identity reads
$$p_3=e_1(e_1^2-2e_2)-e_2e_1+3e_3=e_1^3-3e_1e_2+3e_3,$$
which is the classical expression of $p_3$ in terms of $e_1,e_2,e_3$.

General form of Newton's identities. For $k\ge1$ and $k\le n$,
$$p_k=e_1p_{k-1}-e_2p_{k-2}+\cdots+(-1)^{k-1}e_{k-1}p_1+(-1)^k\,k\,e_k,$$
and for $k>n$ the last term disappears. For $n=3$:
$$p_1=e_1,\qquad p_2=e_1p_1-2e_2,\qquad p_3=e_1p_2-e_2p_1+3e_3.$$

Conclusion. $p_3=e_1p_2-e_2p_1+3e_3$; equivalently $p_3-e_1p_2+e_2p_1-3e_3=0$.