find $x$ where $x^3$ $\equiv$ $1\mod p$, given a primitive root modulo of $p$ How to efficiently find $x$ where $x^3$ $\equiv$ $1\mod p$, given a primitive root of $p$ in the range of $[0,p-1]$?
$p$ is a prime.
Is is true that it only has three roots?
So if one of the primitive roots is given, how to efficient found all such x?

COMMENT (posted because is long for write as a comment).- There are three roots if and only if $p=3m+1$ and only a root if and only if $p=3m+2$.
Since $x^{p-1}-1=0$, if $x^3-1=0$ has three roots then
$\dfrac{x^{p-1}-1}{x^3-1}\in\mathbb F_p$ and if only one root (obviously equal to $1$) then $\dfrac{x^{p-1}-1}{x^3-1}\notin\mathbb F_p$.
Examples.-In $\mathbb F_{13}$ one has $\dfrac{x^{12}-1}{x^3-1}=(x^6+1)(x^3+1)$ and $x=1,3,9$
In $\mathbb F_{11}$ one has $\dfrac{x^{10}-1}{x^3-1}=x^7+x^4+x+\dfrac{1}{x^2+x+1}\notin\mathbb F_{11}[x]$ and just $x=1$
