Relationship between a determinant and the roots of a cubic equation

The $S_j$ obey the recurrence $$S_{n+3}+pS_{n+1}+qS_n=0.$$ But $S_0=3$, $S_1=0$, $$S_2=(a+b+c)^2-2(ab+ac+bc)=-2p.$$ So $$S_3=-pS_1-qS_0=-3q$$ and $$S_4=-pS_2-qS_1=2p^2$$ etc.

$\Delta=VV^T$ where $$V=\begin{pmatrix}1&1&1\\ a&b&c\\ a^2&b^2&c^2 \end{pmatrix}$$ is a Vandermonde matrix. Then $\det V=(c-b)(c-a)(b-a)$, so $\det\Delta=0$ iff two of the roots are equal. If they are all real and distinct, $\det V$ is real, and $ \det\Delta =(\det V)^2>0$. If two are imaginary, they are complex conjugates, and $\det V$ is purely imaginary, and $\det\Delta <0$.

We have, by Vieta's formulas, $$ e_1=0, e_2=p, e_3=-q, e_4=\dots=0 $$ So using Newton-Girard formulae: \begin{align*} S_1 &= e_1 = 0\\ S_2 &= e_1S_1 - 2 e_2 = -2p\\ S_3 &= e_1S_2 - e_2S_1 + 3e_3 =-3q\\ S_4 &= e_1S_3 - e_2S_2 + e_3S_1 +4e_4=2p^2 \end{align*} and now feed in your determinant for question (1).

For question (2), note that $\Delta$ is the discriminant $(a-b)^2(b-c)^2(c-a)^2=-4p^3-27q^2$. It is plausible that solving the depressed cubic was examinable in 1947, certainly it is still covered by some high school textbook today.