I have to find a splitting field of $x^{6}-3$ over $\mathbb{F}_{7}$

I want to find a splitting field of $x^{6}-3$ over $\mathbb{F}_{7}$. I learned that Finite field containing $\mathbb{F}_{7}$ is the form of $\mathbb{F}_{7^m}$ and it is normal extension. So I've tried to find smallest m containing a single root of $x^{6}-3$. If $x^{6}-3$ is irreducible in $\mathbb{F}_{7}$, then $\mathbb{F}_{7^m}$ is splitting field of $x^{6}-3$. But I don't know where I should start. Even, I cannot prove that $x^{6}-3$ is irreducible. For polynomial of degree less than 3, there is a method to determine whether it is irreducible or not. But this is not the case. And the try using Gauss' Lemma and Eisenstein is failed, because I cannot find ring $R$ of which field fraction is $\mathbb{F}_{7}$. Is it wrong approach for this kind of question?

$x^6-3$ has no roots in $\mathbb{F}_7$, since $3$ is not a quadratic residue $\!\!\pmod{7}$. Moreover, in $\mathbb{F}_7[x]$: $$ \gcd\left(x^6-3,x^{49}-x\right) = \gcd\left(x^6-3,x^{48}-1\right) = \gcd\left(x^6-3,3^8-1\right) = 1 $$ hence there is no quadratic irreducible polynomial over $\mathbb{F}_7$ that is a divisor of $x^6-3$.

In a similar way: $$ \gcd\left(x^6-3,x^{343}-x\right) = \gcd\left(x^6-3,x^{342}-1\right) = \gcd\left(x^6-3,-1\right) = 1 $$ hence there is no irreducible polynomial over $\mathbb{F}_7$ that divides $x^6-3$, hence $x^6-3$ is an irreducible polynomial over $\mathbb{F}_7$ and its splitting field is isomorphic to $\mathbb{F}_{7^6}$.

Yet another approach, which works because of the special nature of your polynomial $X^6-3$. Notice first that $3$ is a primitive element of $\Bbb F_7$, that is, it generates the cyclic group of order six consisting of the nonzero elements of this field. So any $\lambda$ with $\lambda^6=3$ must be a primitive $36$-th root of unity in its field.

Now you want the smallest extension of $\Bbb F_7$ having thirty-sixth roots of unity, in other words, you want the smallest $m$ such that $36|(7^m-1)$, in other words, smallest $m$ with $7^m\equiv1\pmod{36}$, in other words, you want the period of $7$ in the group $(\Bbb Z/36\Bbb Z)^*$. Well, the powers of $7$ in $\Bbb Z/36\Bbb Z$ are $1,7,13,19,25,31,1,\cdots$, so $m=6$ here.