 Solution

The cubic construction is a formal interior solution, not an isolated stellar equilibrium. The problem is more basic than whether a cubic star would be stable: the cubic polytrope fails isolated gravitational matching.

For a direct corner-force obstruction for a cubic polytrope, examine the corner at the origin. Every first derivative of the sine-product mass density vanishes there, so $\Phi =C-2K\rho$ predicts $\nabla \Phi (0)=0$. But the actual Newtonian gravitational potential of a positive mass density confined to this cube hasand similarly for the other two components. The integral is finite and strictly negative because the mass density is positive throughout the interior. The physical gravitational field is continuous up to this boundary point, so it cannot match the zero gradient of the proposed interior potential. An additive constant in the potential cannot fix that disagreement. The counterexample to a global interpretation is therefore the very constructed mass density: it solves the local equations but not the global gravitational boundary problem.

A nonrotating isolated fluid supported by an isotropic barotropic pressure normally has a spherical equilibrium. Rotation, tides and magnetic stresses can produce smooth departures from sphericity, but none of these is included in the cubic model, and no rigid walls exist to support its faces and corners. Observationally, ordinary stellar photospheres are consistent with approximately spherical shapes, or with oblateness and tidal distortions where those effects matter; there is no basis for expecting sharply bounded cubic stars. Thus the theoretical obstruction and the expected observations agree.