It can be proven that
If $$P(x)=ax^3+bx^2+cx+d$$ has two roots being opposite of each other i.e. $\alpha$ and $-\alpha$, then $$ad=bc$$
This instantly made me think of a zero determinant of the general $2\times2$ matrix. I cannot think of any obvious reason as to why this might be the case. Is this really just a coincidence, or is there more to it than this?
On
If $\alpha=0$ then we must have $c=d=0$ so the result is obvious. If $\alpha\neq 0$, note that the matrix $$ \begin{bmatrix} a&b\\ c&d \end{bmatrix} $$ sends both $\begin{bmatrix}\pm\alpha\\1\end{bmatrix}$ (which are linearly independent) to the subspace $\left\langle\begin{bmatrix}1\\-\alpha^2\end{bmatrix}\right\rangle$. Hence it must have zero determinant, i.e., $ad-bc=0$.
Let $\beta$ be the third root. Then\begin{align}P(x)&=a(x-\alpha)(x+\alpha)(x-\beta)\\&=a(x^3-\beta x^2-\alpha^2x+\alpha^2\beta)\end{align}and therefore…