Suppose that $p > 3$ is prime, and suppose that $r$ is a primitive root mod $p$. Prove that if $p ≡ 1$ mod $3$ then $r$ is a cubic non-residue mod $p$.
I have tried to manipulate the Euler Criterion and Legendre symbols, but I cannot get to a point where I can introduce the primitive root $r$.
On
On a finite abelian group of order $n$, the map $x \mapsto x^3$ is a homomorphism; it is a bijection iff $\gcd(3,n)=1$. This is proved easily by considering the kernel. Since $3$ is prime, $\gcd(3,n)=1$ iff $n \not \equiv 0 \bmod 3$.
In our case, $n=p-1$. If $p \equiv 1 \bmod 3$, then $\gcd(3,n)=3 > 1$ and the map $x \mapsto x^3$ is not surjective. Therefore, $r$ is not in the image.
HINT: If $p\equiv1\pmod{3}$ then $(\Bbb{Z}/p\Bbb{Z})^{\times}$ is cyclic of order $p-1$, which is a multiple of $3$.