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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?

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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 $x = y = z = \frac {\sqrt3} 3$.

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    Thank you @Théophile. Is the volume of the sphere $8xyz$? If so is it because there are 8 points that touch the sphere, and does that require that the cuboid be a cube and not rectangular, or does that distinction not matter? Let's say for example that the sphere accommodates a cuboid that's 1x3x5. Would that mean the volume of the cuboid is 15 cubic units, and the sphere is 120 cubic units? My knowledge of basic geometry is ... basic. – 2017-01-04
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    @mkinson Glad to help. A sphere of radius has volume $\frac43 \pi r^3$, as you pointed out in your original question, and a cuboid has volume $lwh$. In this case, if one of the corners of the cuboid is $(x,y,z)$, then it has other corners at $(\pm x, \pm y, \pm z)$. So the side lengths are $l=2x, w=2y, h=2z$, and the total volume is $lwh=(2x)(2y)(2z)=8xyz$. Compare to the two-dimensional analogy of a rectangle inscribed in a circle. If one of the corners is $(x,y)$, then the area of the rectangle is $4xy$. – 2017-01-04
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    As for your example, I think you're mixing up the cuboid and the sphere. If the inscribed cuboid is $1 \times 3 \times 5$, then its volume is indeed $15$. In this case, the corners are at $(\pm \frac12, \pm \frac32, \pm \frac52)$. The radius of the sphere, using Pythagoras, is $r = \sqrt{\left(\frac12\right)^2 + \left(\frac32\right)^2 + \left(\frac52\right)^2} = \frac12 \sqrt{35}$, and the volume of the sphere can then be found with $\frac43 \pi r^3$. – 2017-01-04
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    Thank you again, that helps immensely! – 2017-01-04
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    One last thing: you'll find that your problem in fact has no solution, because even the largest cuboid inscribed in the sphere falls quite short of half the volume of the sphere... – 2017-01-04