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?

Assume that you have an irreducible polynomial $P$ of degree $d$ dividing your polynomial. Then taking $r$ to be the class of $X$ in the field :
$$\mathbb{F}_{7^d}:=\frac{\mathbb{F}_7[X]}{(P)} $$
We know that $r^6=3$ in this new field  (since $P$ divides $X^6-3$). On the other (because you know Lagrange's theorem) you have that :
$$r^{7^d-1}=1 $$
Now I claim that both equations are not compatible when $d=2$ or $3$. Let us do the case $d=2$. In that case we have $r^{48}=1$ and $r^6=3$ but $48=6.8$ so : $r^{48}=3^8=3$ mod $7$ which is not $1$. 
And the case $d=3$ $r^{342}=1$ and $r^6=3$ but $342=6.57$ so that $r^{342}=3^{57}=3$ mod $7$ which is not $1$. 
Hence you cannot have an irreducible polynomial of degree $1$, $2$ or $3$ dividing $X^6-3$, this means that your polynomial is irreducible. 
