= Solution
By <Vieta's formulas>, the elementary <symmetric polynomials> in the roots give
$$
c_1=-(\alpha_1+\alpha_2+\alpha_3),\qquad
c_2=\alpha_1\alpha_2+\alpha_2\alpha_3+\alpha_3\alpha_1,\qquad
c_3=-\alpha_1\alpha_2\alpha_3.
$$
For each root, $\alpha_j^3+c_1\alpha_j^2+c_2\alpha_j+c_3=0$. Multiplying by $\alpha_j^k$ and summing gives the <power sum> recurrence
$$
\boxed{s_{k+3}+c_1s_{k+2}+c_2s_{k+1}+c_3s_k=0\qquad(k\ge0).}
$$
Its initial values are $s_0=3$, $s_1=-c_1$ and $s_2=c_1^2-2c_2$. They are <polynomials> with <integer> coefficients in $c_1,c_2,c_3$. The recurrence uses only addition and multiplication by those coefficients, so <induction> proves the same for every $s_k$. This is the cubic case of <Newton's identities>.
For $c_1=0$, put $\alpha_1=u$, $\alpha_2=v$, $\alpha_3=-u-v$. Then $c_2=-(u^2+uv+v^2)$ and $c_3=uv(u+v)$. Expanding the product of root differences gives
$$
\begin{aligned}
\Delta&=(u-v)^2(u+2v)^2(2u+v)^2\\
&=4(u^2+uv+v^2)^3-27u^2v^2(u+v)^2
=\boxed{-4c_2^3-27c_3^2}.
\end{aligned}
$$
This is the <discriminant of a depressed cubic> in depressed form, and the displayed identity also proves its expression in integer-coefficient <symmetric polynomials>.
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