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For the cubic polytropic interior, $\nabla \rho$ vanishes at each vertex, so the locally defined potential $\Phi =C-2K\rho$ predicts zero gravitational acceleration there. But at the vertex $(0,0,0)$, the actual self-gravitational acceleration of the positive mass distribution iswhose three components are strictly positive. The fields cannot match continuously. Thus the interior Helmholtz equation and zero density on the faces do not construct an isolated static cubic star. External stresses or an external gravitational field would be needed to realize such a boundary-value construction.

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.