 Solution

For $P=K{\rho }^2$, the specific enthalpy isHydrostatic equilibrium says $\nabla (H+\Phi )=0$. Taking a Laplacian and using the gravitational Poisson equation gives the Helmholtz equation

For a spherical star, regularity at the centre selects the $n=1$ stellar polytropeIts first zero is $kR=\pi$, henceDirect integration gives $M=4{\rho }_c{R}^3/\pi$, and therefore

On the cube, the separated positive solutionvanishes on all six faces. It solves the same Helmholtz equation whenThe mean of each sine over $[0,L]$ is $2/\pi$, so

The interior fields formally solve the local structure equations, but an isolated fluid surface must be an equipotential and its interior gravitational field must match a decaying exterior solution with continuous normal derivative. A cube does not satisfy the global free-boundary conditions for a nonrotating self-gravitating barotrope. Such sharp planar faces and edges are also not observed in stars; ordinary pressure and gravity smooth the body toward a sphere.