Is $x^6-3$ irreducible over $\mathbb{F}_7$? I know that $\mathbb{F}_7=\mathbb{Z}_7$, and the all possible solutions of $x^6-1=0$ over $\mathbb{Z}_7$ are 1~6, so if we let the root of equation $x^6-3$ as $t $ then the solutions of $x^6-3=0$ is $t$~$6t$, which shows that all the solutions are not in $\mathbb{Z}_7$. However I can't say that $x^6-3$ is irreducible by this fact(since there could exist 2 or more degree polynomial divides $x^6-3$.) Is it irreducible over $\mathbb{Z_7}$? Why?

Same question here. I am new on this site. I dont know how to share same question link. (as a comment or answer). But I wanna inform every one that there is same question.
and this was my answer in the other question:
I wanna use an easy tool: 
1) it is not neccesary to check $3$ is square of not in $\mathbb F_7$. $x^6-1 \equiv 0 \mod 7$ for all $x \in \mathbb F_7^*$ which is well-known.
2) Use same trick again $3^6 \equiv 1 \mod 7$. That means $x^{36}\equiv 1 \mod 7$. Now $36\mid(7^6-1)$ trivially, it is $(7-1)(7^5+\cdots+1)$.  Moreover, $36 \not \mid (7^n-1)$ for $n <6$.  This means it is irreducible.
