Cut a cuboid from the center of a sphere so that the cuboid and sphere have equal volume I'm trying to determine if it is possible for a cuboid to be removed from a sphere so that the volume of the cuboid and the volume of the remaining portions of the sphere are equal, where the corners of the cuboid are points along the sphere's circumference. I realize that the volume of a cuboid is simply $L*W*H$ and the volume of a sphere is $(4/3)πr^3$ but I don't think I can use $L*W*H$ = $(4/3)πr^3$ - $L*W*H$ since it doesn't necessarily follow the above listed constraints.. Is there another way to write out this equation so that I can account for the cuboid x, y, z points being dependent on the sphere's circumference?

Let's assume that the radius of the sphere is $1$. Then we want to find a cuboid with volume $\frac12 \left(\frac43 \pi \right) = \frac23 \pi$.

If one corner of the cuboid is $(x,y,z)$, then its volume is $8xyz$, so we have $xyz = \frac18 \left(\frac23 \pi \right) = \frac1{12}\pi$. We also have $x^2+y^2+z^2=1$, since the corner lies on the surface of the sphere. So we have:

$$xyz = \frac1{12}\pi\\\ x^2 + y^2 + z^2 = 1$$

A good place to start would be to check how big the largest possible cuboid is. This occurs at $x = y = z = \frac {\sqrt3} 3$.