Is the root of $x^3+6x+1$ collapsible? (any collapsing has degree $\ge 32$)

Is the root of $x^3+6x+1$ collapsible?

Call $\alpha$ collapsible if there is some $f\in\mathbb{Q}[x]$ of degree $\ge 1$, with nonzero rational leading coefficient and all roots rational, such that $f(\alpha)\in\mathbb{Q}$. The conjecture behind this mission is that every algebraic number is collapsible; the first open case is the cubic one. This theorem is the smallest cubic still resisting: no collapsing of a root of $x^3+6x+1$ is known.

What is already established. This cubic has discriminant $\Delta=-4\cdot 6^3-27=-891<0$. Writing $x^3+6x+1=(x+2u)\big((x-u)^2+v^2\big)$ gives $u\approx 0.0829528$, $v\approx 2.4537000$, so with $\psi=\arctan(3u/v)\approx 0.10108$ we get $\pi/\psi\approx 31.08$. A machine-checked archimedean bound (CollapsibleCubics.pi_div_psi_le_natDegree, a proved milestone of this mission) then forces

for every collapsing $f$ of this root. So this is not a gap that low-degree search can close: exhaustive checks at degree $4$ (up to height $800$, denominator $40$) and extensive MILP searches found nothing, and provably could not have. Progress means exhibiting a witness of degree $\ge 32$, or proving none exists.

Either resolution is significant. A collapsing would settle the smallest open case. A disproof — showing this root is not collapsible — would refute the general conjecture outright, so this statement is a genuine two-sided target and should not be assumed true.

The statement is given for $\alpha$ in an arbitrary field $K$ containing $\mathbb{Q}$, so it covers the root in $\mathbb{R}$, in $\mathbb{C}$, and in $\mathbb{Q}[x]/(x^3+6x+1)$ alike. Over $\mathbb{Q}$ itself the hypothesis is unsatisfiable (the cubic has no rational root), so the content is exactly the extension case.