With edge and compass construction, given cubes of volumes $a^3,b^3$, can one construct a cube of volume $a^3+b^3$? Suppose one has cubes A and B of volumes $r_A^3$ and $r_B^3$.  Using only ruler and compass constructions, I need to determine whether it is possible to construct a cube of volume $r_A^3+r_B^3$.  I have already shown it is possible to take given squares of two areas and construct the square with area the sum of their areas--I did this by arranging the squares at a right angle to each other at one vertex, and from there using the Pythagorean Theorem, the rest is simple.  In this setting, even if there is some 3D version of the Pythagorean Theorem, I doubt I can use it in this problem.
If it's not possible I could try to argue that, if one could do this, then one could do some other construction which we know is impossible.  For instance we know that it is impossible to square the circle or trisect the angle.  Squaring the circle kind of sounds relevant to this because we are dealing with spheres.  But I haven't been able to see a useful connection beyond that.  Trisecting the angle might also apply because we are now in three dimensions and perhaps that bears on this somehow, like maybe the line through a vertex and midpoint of an opposite face of the cube is 1/3 the angle or something ... I don't know, at this point I'm just brain-storming.
 A: This is a more general case of a classic problem from antiquity known as doubling the cube. This was proven to be impossible in 1837. The desired cube cannot be constructed, as it requires constructing cube roots of integers, which is not possible in general.
A: There are no solutions with natural numbers, by Fermat's Last Theorem, but certain other combinations of constructible numbers do admit solutions. One such combination is
$(\sqrt{85}-1)^3+8^3=(\sqrt{85}+1)^3.$
If we expand the cubes of the parenthesized sum and difference with the Binomial Theorem, we find that the difference is $2×(3×85+1)=512=8^3$ as claimed.
We may compare this relation with the Pythagorean triple having a leg of $8$ and the other two sides differing by $2$:
$15^2+8^2=17^2$
$(\sqrt{256}-1)^2+8^2=(\sqrt{256}+1)^2.$
The $a$ and $c$ sides are shorter when the exponents are $3$ rather than $2$, but still greater than the fixed $b$ side $8$, and since $c-a$ is also held constant the first triangle must have more nearly equal angles than the second. Therefore with the second triangle being a right triangle the first triangle must be acute. This is a general property of triangles satisfying $a^3+b^3=c^3$.
