Relationship between a determinant and the roots of a cubic equation The following is a question in a 1947 university entrance exam.

Let $a$, $b$ and $c$ be the roots of cubic equation:
$$ x^3+px+q=0$$
and denote
$$ S_n=a^n+b^n+c^n$$
Question (1): express the below determinant in terms of $p$ and $q$:
$$ \Delta = \begin{vmatrix} 
S_0 & S_1 & S_2 \\
S_1 & S_2 & S_3 \\
S_2 & S_3 & S_4
\end{vmatrix} $$
Question (2): show that

*

*When $\Delta>0$, $a$, $b$ and $c$ are distinct real roots

*When $\Delta<0$, one root is real, and the other two are complex conjugated

*When $\Delta=0$, all roots are real, and at least two of them are equal



For me, I can just plug in the cubic formula (such as using Mathematica) to get the answer to question (1). However such computer tools are not available in 1947. Also considering that it is in an exam environment, one is not expected to spend too much time on it. So my question is, whether there is an ingenious way to solve it?
 A: 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.
A: 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 $\det\Delta <0$.
