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Choose zero gravitational potential at infinity. Apply superposition to a uniform sphere of positive density and the two voids represented by spheres of negative density. The shell theorem makes every component act as a point mass outside the large sphere. With $c_2=(1/2,0,0)$ and $c_3=(-1/4,0,0)$,The volume factors are the cubes of the three radii.

Inside the smaller void, the large sphere and the removed small sphere have interior fields $-Cx$ and $+C(x-c_3)$, where $C=4\pi G\rho /3$. They combine into the uniform gravitational field in an off-center spherical cavity, $-C{c}_3=C(1/4,0,0)$. The point is outside the other void's sphere, whose removed mass contributes a repulsive field. ThusAn equilibrium must lie on the line through the two centers. Write $x=q{e}_1$ with $-1/2<q<0$. Its equation is $1/4-1/[8(1/2-q{)}^2]=0$, givingThis point is inside the smaller void, but it is not stable. Let $d=|x_{\ast }-c_2|=1/\sqrt{2}$. The field's Jacobian matrix has eigenvaluesThe axial displacement is restoring, while transverse displacements accelerate away from equilibrium. Therefore there is no point in that void where a test particle remains stably at rest. The cavity equilibrium destabilized by a second void is a saddle; existence of a zero force is insufficient for stability.