prove or disprove:
Suppose that p is odd prime such that $p\equiv 1\mod{3}$ and a is primitive root mod $p$. Let $a=r^{\frac{p-1}{3}}$ then $1$, $a$, $a^2$ are solution to $x^3\equiv 1\mod{p}$ that are distinct $\mod{p}$.
I guess it is true statement
since $p$ is odd prime such that $p\equiv 1\mod{3}$ then $p-1+3k$ also, since a is primitive root $\mod{p}$ then $(a,p)=1$
Let $a=r^{\frac{p-1}{3}}$ so let $a^3=r^{p-1}$
any help with that please how can I conclude that $1$, $a$, $a^2$ are solution to $x^3\equiv1\mod{p}$ that are distinct $\mod{p}$?