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The sine-product cubic polytropic interior has zero mass density gradient at every corner. Its proposed hydrostatic gravitational potential ${\Phi }_{\mathrm{int}}=C-2K\rho$ therefore satisfies $\nabla {\Phi }_{\mathrm{int}}(0)=0$ at the corner $(0,0,0)$. For the actual Newtonian gravitational potential of that positive mass density in $[0,L{]}^3$,Every component is strictly negative: all source positions have positive coordinates and positive interior mass density. The gravitational acceleration $\mathbf{g}=-\nabla {\Phi }_N$ consequently points into the cube, while the potential gradient points in the opposite direction. Neither vector vanishes.

This sign distinction preserves the matching obstruction: the hydrostatic interior predicts zero corner force, but its own mass produces a nonzero corner force. Satisfying the interior Poisson equation for Newtonian gravity and zero density on the faces does not ensure a globally isolated equilibrium. An added external gravitational field or external stresses would change the boundary problem.

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.

A nonspherical configuration requires the vector form of hydrostatic equilibrium and the Poisson equation, rather than the spherical mass-coordinate equations:For the same index-one polytrope, hydrostatic balance inside the positive-density region gives $\Phi =C-2K\rho$. Thus its interior mass density must obey the Helmholtz equationA positive separated solution with the required boundary values isIts maximum is ${\rho }_c$ at the centre of the cube and it vanishes on every face. Its Laplacian is $-3{\pi }^2\rho /L^2$, so it satisfies the interior equations whenTogether with $P=K{\rho }^2$ and $\Phi =C-2K\rho$, this explicitly constructs the cubic polytropic interior.

The mass integral separates into three elementary sine integrals:ThereforeThis construction satisfies the equations inside the prescribed cube. It does not by itself establish the existence of an isolated self-gravitating star: the interior potential must also match the exterior vacuum field generated by that very mass density. That additional physical requirement is addressed in part (c).

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.