Doubling the cube with unit sticks

Who, me? You think I can beat your nice construction within a matter of minutes? Okay, one $2^{1/3}$ polynomial we can look at is $1 - 3 x + 3 x^2 + x^3$, which has algebraic discriminant -108. If we can find braceable points that work nicely with that space, then getting back to $2^{1/3}$ should be easy. Here's some polynomials to extract complex roots from.

$$(x, 1 - 2 x + 2 x^2, 1 + 2 x + 2 x^2, 2 - 2 x + x^2 - 2 x^3 + 2 x^4, 2 + 2 x + x^2 + 2 x^3 + 2 x^4, 1 + 2 x + 6 x^2 - 16 x^3 + 8 x^4, 1 - 6 x + 14 x^2 - 8 x^3 + 8 x^4, 1 + 6 x + 14 x^2 + 8 x^3 + 8 x^4, 1 - 2 x + 6 x^2 + 16 x^3 + 8 x^4)$$

Find the points with unit distances and clean up, and you're left with the following unit-distance graph:

Is that rigid, or does it need more work? What is the chromatic number? Is it easy to get back to $2^{1/3}$? I don't know. But it's a nice graph within the correct algebraic space. See if it helps.

Khodulyov's bracings of regular polygons use a "folded" and very economical Peaucellier–Lipkin linkage that replicates the jack's functionality and then some. In the diagram below the long edges are of length $\sqrt3$ (five sticks in the actual realisation) and not only is $\angle P_0MO=\angle MOM'$ – $C$ is constrained to be on the perpendicular to $MO$ through $M$ and $P_0$ is constrained to be collinear with $P_1M$.

Combined with the question's strategy this gives a $43$-edge bracing of $\sqrt[3]2$ (the $\sqrt3$ diamonds in the second and third images below have been replaced with single edges for clarity):