Let $a,b,c$ be all the roots of $ x^3 + sx + t$. What is the determinant of the matrix $\begin{bmatrix} a & b & c\\ b & c & a\\ c & a & b \end{bmatrix}$?
I wrote the the system for $ p(a)=0,p(b)=0, p(c)=0$, tried to eliminate parameters $s,t$ and substitute into the expression for the determinant $3abc - a^3 - b^3 - c^3$, hoping to get zero, which didn't work. Also, I tried to show that singularity of the matrix of the system $\begin{bmatrix} 1 & a & a^3\\ 1 & b & b^3\\ 1 & c & c^3 \end{bmatrix} * \begin{bmatrix} q\\ p\\ 1 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$ implies that the determinant of the given matrix is zero, which also didn't work. What am I missing?
Should I use the symmetry of the matrix somehow?
Note that $a^3+sa+t=b^3+sb+t=c^3+sc+t=0$ so that (adding the three expressions) $$(a^3+b^3+c^3)+s(a+b+c)+3t=0$$
Now equating coefficients in $x^3+sx+t=(x-a)(x-b)(x-c)$ we get $a+b+c=0$ (coefficient of $x^2$) and $t=-abc$
Substituting these into the previous equation we get: $$a^3+b^3+c^3-3abc=0$$
NOTE: suppose our equation had been $x^3+rx^2+sx+t=0$. We add the three expressions for $a,b,c$ as before to get:
$$(a^3+b^3+c^3)+r(a^2+b^2+c^2)+s(a+b+c)+3t=0$$
Equating coefficients as we did before $r=-(a+b+c), s=(ab+bc+ac), t=-abc$. If we then take the $r$ and $s$ terms to the other side, and substitute the $a,b,c$ expressions we get
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2)-(ab+bc+ac)(a+b+c)=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$
The two middle terms of the cubic each yield factor $(a+b+c)$. This identity is particularly useful when $a+b+c=0$ (as is true here). In that case, the method I used above also works nicely.