I recently investigated $\textrm{Discriminant}_u(\omega P(u) +(z-u)P'(u))$, where $P(u) := u^3 + au + b$ and $\omega$ is a real parameter (with $\omega\in(0,1)\cup(1,3)$) associated with the order of a fractional derivative. In doing so, I came across the following expression (plus some additional terms) that looks quite striking, but I haven't been able to place it:

$$\frac{3a(b-z^3)}{(\omega-1)\,\lambda} + \frac{z}{\lambda^3}\left(a^2\lambda\left(1-\frac{1}{(\omega-1)^2}\right)-3bz\right)$$

(Here $\lambda:= \sqrt[3]{\omega-3}.$)

Due to a natural change of variables used to arrive at the above expression (which reduces the additional terms to the discriminant of the polynomial below), the expression should possibly be thought of as the “image” of the polynomial

$$\widetilde{P}(u) := u^3 + \frac{a}{(\omega-1)\sqrt[3]{\omega-3}}u + \frac{b}{\omega(\omega-3)}$$

under some discriminant-like “function”.

Does the expression above look familiar, or is it perhaps some determinant involving the coefficients in $\widetilde{P}(u)$, and $z$?

As a final note, the above expression is a transformed version of the polynomial $3az^3 + 3\omega z^2 - a^2 (\omega(\omega - 2) - 2)z - 3ab\omega (\omega - 3)$, which is a factor in the non-constant part of $\textrm{Discriminant}_u(\omega P(u) +(z-u)P'(u))$. Perhaps it's easier to attack that polynomial instead, but the reduction of the additional terms into the discriminant of $\widetilde{P}(u)$ under the transformation makes me suspect otherwise.