p is an odd prime, and p does not divide u prove that if p is congruent to 2 mod 3, then every unit is a cubic residue mod p. I don't know where to start with this.
Hi,
If you know a few facts, this is easy. Otherwise, it might be kinda hard.
The multiplicative group of $\displaystyle \mathbb Z_p$ is cyclic with generator, say x (a primitive root mod p). Then the order of $\displaystyle x^3$ is the number of elements in the set $\displaystyle \{x^{3k}\,:\, k\in\mathbb Z\}={p-1\over gcd(p-1,3)}$. Now if 3 divides p - 1, then p is congruent to 1 mod p, but p is congruent to 2 mod 3. So gcd(p-1,3)=1. That is, the order of $\displaystyle x^3$ is p - 1 and so every non-zero element of $\displaystyle \mathbb Z_p$ is a cube; i.e. a cubic residue.