= Solution
For $P=K\rho^2$, the specific enthalpy is
$$
H(\rho)=\int_0^\rho\frac{dP}{\rho'}=2K\rho.
$$
Hydrostatic equilibrium says $\nabla(H+\Phi)=0$. Taking a Laplacian and using the gravitational <Poisson equation> gives the <Helmholtz equation>
$$
\boxed{\nabla^2\rho+k^2\rho=0},
\qquad
k^2=\frac{2\pi G}{K}.
$$
For a spherical star, regularity at the centre selects the $n=1$ <stellar polytrope>
$$
\rho(r)=\rho_c\frac{\sin kr}{kr}.
$$
Its first zero is $kR=\pi$, hence
$$
\boxed{R=\frac\pi k=\left(\frac{K\pi}{2G}\right)^{1/2}}.
$$
Direct integration gives $M=4\rho_cR^3/\pi$, and therefore
$$
\boxed{\frac{\bar\rho}{\rho_c}
=\frac{3M}{4\pi R^3\rho_c}=\frac3{\pi^2}}.
$$
On the cube, the separated positive solution
$$
\rho(x,y,z)=\rho_c
\sin\frac{\pi x}{L}
\sin\frac{\pi y}{L}
\sin\frac{\pi z}{L}
$$
vanishes on all six faces. It solves the same Helmholtz equation when
$$
\frac{3\pi^2}{L^2}=k^2,
\qquad
L=\left(\frac{3\pi K}{2G}\right)^{1/2}.
$$
The mean of each sine over $[0,L]$ is $2/\pi$, so
$$
\boxed{\frac{\bar\rho}{\rho_c}=\left(\frac2\pi\right)^3=\frac8{\pi^3}}.
$$
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.
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