What about zero? Can zero be the sum of three cubes? Actually, I read that somebody proved it can't be done, but the proof is too long to fit in this comment. ;-)
For this particular problem, none of the cubes are allowed to be 0
I don't think that's a restriction that's usually enforced in this case. It seems nicer to allow the cubes to be zero since then the conjecture that the achievable sums are everything not 4 or 5 mod 9 can be stated without special cases.
Yes. The joke the original commenter is playing is that Fermat's `x³ + y³ = z³` is equivalent to `x³ + y³ + (-z)³ = 0`.