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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.

The same polytrope of index one interior equation is ${\nabla }^2\rho +(2\pi G/K)\rho =0$. A positive separated solution with zero density on all six cube faces is the cubic polytropic interiorIts Laplacian is $-3{\pi }^2\rho /L^2$, so the required side length isWith $\Phi =C-2K\rho$ and $P=K{\rho }^2$, it satisfies the interior hydrostatic equilibrium and Poisson equation for Newtonian gravity. Integrating each sine factor yieldsThis formal interior solution is not an isolated physical cubic star. The cubic polytrope fails isolated gravitational matching: at a vertex, the product of sines has $\nabla \rho =0$, so the interior potential predicts zero gravitational acceleration. At the vertex $(0,0,0)$, however, the field generated by its own positive mass isand each component is strictly positive. There can be no continuous matching to the isolated external field without additional forces or mass sources. Thus solving the interior density equation and imposing zero face values is insufficient. Fluid stars also have no rigid structure to maintain sharp cubic faces, and observed stellar shapes are approximately spherical or rotationally flattened rather than cubic.