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So far, I have split the primes up into two cases:

For p = 3k+2, every a(mod p) has a cube root.

For p = 3k+1, I don't know which a it is true for, but I did a few examples and noticed a couple of things:

* If a is a cube, so is -a

* If a is a cube, it has exactly 3 cube roots

Does this have anything to do with p-1= 0(mod3)?