I think it will be helpful to organise the discussion in the comments into something more coherent.
In my opinion, the biggest challenge is less about answering the question at hand than it is about laying out the background theory in a sensible order, taking care to avoid circular logic. For me, the most sensible starting point is to characterise the fields of the form $\mathbb F_{p^n}$, where $p$ is a prime. As mentioned in the comments, the question you asked becomes trivial once we've fully characterised $\mathbb F_{p^n}$.
Proposition 1: Let $L$ be a splitting field of $x^{p^n} - x$ over $\mathbb F_{p}$. Then $L$ contains $p^n$ elements.
This proves the existence of a field with $p^n$ elements.
Sketch proof of Proposition 1:
- Let $E: = \{ \alpha \in L : \alpha \text{ is a root of } x^{p^n} - x \}$. Then $E$ contains precisely $p^n$ elements. (If it were the case that $x^{p^n} - x$ has repeated roots, then $x^{p^n} - x$ and its derivative would have a common factor. But this is impossible, since the derivative of $x^{p^n} - x$ in a field of characteristic $p$ is $-1$.)
- $E$ is a field. (A consequence of the algebraic identities $(x + y)^{p^n} = x^{p^n} + y^{p^n}$ and $(xy)^{p^n} = x^{p^n}y^{p^n}$ in fields of characteristic $p$.)
- $L$ is a splitting field for $x^{p^n} - x$, so $x^{p^n} - x$ does not split completely in any proper subfield of $L$. But $E$ is a subfield of $L$ in which $x^{p^n} - x$ splits completely. Therefore, $L = E$.
Proposition 2: Let $L$ be a field with $p^n$ elements. Then $L$ is a splitting field of $x^{p^n} - x$ over $\mathbb F_p$.
Since splitting fields are unique up to isomorphism, this proves the uniqueness of the field with $p^n$ elements (up to isomorphism).
Sketch proof of Proposition 2: Let $L^\times$ be the multiplicative group of non-zero elements of $L$. By Lagrange's theorem, the order of every element of $L^\times$ divides $p^n - 1$. Therefore, $x^{p^n} - x = 0$ for all $x \in L$.
Definition: We define $\mathbb F_{p^n}$ to be the unique field containing $p^n$ elements.
This definition makes sense, because Proposition 1 proves that such a field exists, and Proposition 2 proves that such a field is unique.
Proposition 2 also tells us that $\mathbb F_{p^n}$ is a splitting field for $x^{p^n} - x$. And that answers the question you posted. The question you posted is a special case of this result, for $p = 2$ and $n = 3$.
However, you went about the problem in a different way. You factorised the polynomial $x^{p^n} - x$ into a product of irreducible factors (in the special case where $p = 2$ and $n = 3$). You then went about determining the splitting fields for each of the irreducible factors of this polynomial, with the intention of using this as a first step towards solving your problem.
Determining the splitting fields for irreducible polynomials in $\mathbb F_p[x]$ seems like a manageable task that we ought be able to accomplish. So I think I should explain how we might go about accomplishing this task.
I've given this more thought than I had at the time of the discussion in the comments, and unfortunately, I can't think of any reasonable way to tackle this that doesn't use Galois theory. This is because I want to use the following lemma, which I'm going to prove with Galois theory.
Lemma: For every $n$ that divides $m$, $\mathbb F_{p^m}$ contains a unique subfield that is isomorphic to $\mathbb F_{p^n}$. $\mathbb F_{p^m}$ contains no other subfields.
Sketch proof of Lemma:
- $\mathbb F_{p^m}$ is a Galois extension of $\mathbb F_p$, since it is the splitting field for the separable polynomial $x^{p^m} - x \in \mathbb F_p[x]$.
- $Gal(\mathbb F_{p^m}, \mathbb F_p)$ is a group of order $m$.
- $Gal(\mathbb F_{p^m}, \mathbb F_p)$ is cyclic: it is generated by the Frobenius automorphism $\sigma : x \mapsto x^p$.
- Since $Gal(\mathbb F_{p^m}, \mathbb F_p)$ is cyclic, it has precisely one subgroup of order $m/n$, for each $n$ that divides $m$, and no other subgroups. By the Galois correspondence, $\mathbb F_{p^m}$ has precisely one subfield isomorphic to $\mathbb F_{p^n}$, for each $n$ that divides $m$, and no other subfields.
Using this lemma, we can prove:
Proposition: Let $f(x)$ be any irreducible polynomial of degree $n$ in $\mathbb F_p[x]$. Then the splitting field of $f(x)$ over $\mathbb F_p$ is $\mathbb F_{p^n}$.
Proof: To prove this, let $L$ the splitting field of $f(x)$. Then $L$ is a finite field of characteristic $p$, so $L$ must be isomorphic to $\mathbb F_{p^m}$, for some $m$. Our task is to prove that $m = n$.
If $\alpha \in L$ is a root of $f(x)$, let $\mathbb F_p(\alpha)$ denote the smallest subfield of $L$ containing $\mathbb F_p$ and $\alpha$. Then $\mathbb F_p(\alpha)$ is an extension of degree $n$ over $\mathbb F_p$, since $\alpha$ is the root of a degree $n$ irreducible polynomial in $\mathbb F_p[x]$. So $\mathbb F_p(\alpha)$ contains $p^n$ elements, i.e. $\mathbb F_p(\alpha)$ is isomorphic to $\mathbb F_{p^n}$.
But our Lemma tells us that $L$ contains a unique subfield $E$ that is isomorphic to $\mathbb F_{p^n}$. Therefore, $E$ is equal to $\mathbb F_p(\alpha)$, for all roots $\alpha$ of $f(x)$. So $f(x)$ splits completely in $E$, but does not split completely in any subfield of $E$. Thus $L = E$, which completes the proof.
Returning to your question, you wanted to find the splitting field for a polynomial of degree $8$ in $\mathbb F_p[x]$ that factorises as a product of two linear factors and two irreducible cubic factors in $\mathbb F_p[x]$. By similar logic to the above, you can prove that the splitting field of any such polynomial is $\mathbb F_{p^3}$.
Indeed, if $L$ is the splitting field for this polynomial of degree $8$, then the splitting field for each of the irreducible cubic factors is a subfield of $L$ that is isomorphic to $\mathbb F_{p^3}$. But $L$ contains a unique subfield $E$ that is isomorphic to $\mathbb F_{p^3}$. So both of the irreducible cubic factors split completely in $E$ (as do the linear factors, trivially). Hence $L = E$.
Best Answer
Since $2^r-1$ is odd, there is no way for it to be a multiple of $12$. While the polynomial $f(X)=X^{12}-1$ has degree $12$, it does not have $12$ distinct roots in its splitting field. Indeed, as $f'(X)=0$, every root of $f$ is a multiple root. And as you found out (in a different way), $(X^3+1)^2=(X^6+1)$ and $(X^6+1)^2=X^{12}+1$ so that $$f(X)=(X^3+1)^4=(X+1)^4(X^2+X+1)^4,$$ i.e., we "really" want to find only the third roots of unity. The rule of thumb you started from holds only for separable extensions (i.e., when all roots are distinct)