4 ms·
Yes. 0³+0³+0³. Or x³+(-x)³+0³.
by 3JPLW 7y ago
Yes. 0³+0³+0³. Or x³+(-x)³+0³.
- FreeFull 7y agoFor this particular problem, none of the cubes are allowed to be 0
- OscarCunningham 7y agoI 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.
- 3JPLW 7y agoYes. The joke the original commenter is playing is that Fermat's `x³ + y³ = z³` is equivalent to `x³ + y³ + (-z)³ = 0`.