Splitting field of $x^{21}-1$ over $\mathbb F_3$ 
Let $\mathbb F_3$ be the field with 3 elements and $\overline{\mathbb F}_3 $ its algebraic closure. Let $K$ be the splitting field of $f(x)=x^{21}-1$ (I guess over $\mathbb F_3$). Find the number of zeros of $f$ in $\overline{\mathbb F}_3 $ and the number of elements in $K$.

There are 3 questions below in different parts of the text, they are in italics.
First of all, I don't understand why would anyone ask about the number of zeros of $f$ in the closure, since it is always equal to the degree of the polynomial. What am I supposed to answer here?
For the second question. Since $\mathbb F_3$ has characteristic 3, $x^{21}-1=(x^7-1)^3$. The set of roots of $f$ is the set of roots of $g=x^7-1$. So the splitting field of $f$ is the same as that of $g$. Now let $a$ be a root of $g$ in the algebraic closure of $\mathbb F_3$. Consider the extension $\mathbb F_3(a)$. Now


*

*either $a=1$ in $\mathbb F_3(a)$ or

*the order of $a$ in the multiplicative group of the extension $\mathbb F_3(a)$ is $7$.


Note that $|\mathbb F_3(a)|=3^n$ since $\mathbb F_3(a)$ is a vector space over $\mathbb F_3$. In the second case above, $3^n-1$ must be divisible by $7$, whence $n\ge 6$. What to do with the first case?
Consider $F=\mathbb F_{3^6}$. There is an element $a$ of order $7$ in the multiplicative group of that field. Then $a^2,\dots,a^6$ also have order $7$. So they all are roots of $g$ (or equivalently $f$). Thus $F$ contains all roots of $f$. How to show it is the minimal extension containing the roots of $f$?
 A: Since in $\mathbb{F}_3$ we have $x^{21}-1=(x^7-1)^3$, $x^{21}-1$ has seven roots in $\overline{\mathbb{F}}_3$, all of them with multiplicity $3$. Indeed $x^7-1 = (x-1)\Phi_7(x)$ and the degree of the splitting field $\mathbb{K}$ of $\Phi_7(x)$ over $\mathbb{F}_3$ is given by the least $d$ such that $7\mid(3^d-1)$, i.e. $d=6$, since $3$ is a generator for $\mathbb{Z}/(7\mathbb{Z})^*$.
$$ \mathbb{K}=\mathbb{F}_3[x]/(\Phi_7(x)) \simeq \mathbb{F}_{3^6}$$
is a finite field with $3^6=729$ elements.
A: As you've said the number of zeroes equals the degree of the polynomial. However I suspect that the question asks for the number of distinct zeroes of maybe the number of zeroes in F.
For the second part note that $1$ is one of the roots. Also there are not other roots in $\mathbb{F}_3$. So let $a$ be such a root in $\overline{\mathbb{F}}_3$. This means that you don't have to split the problem in two cases, as you will have to adjoin $a \not = 1$ to get the splitting field.
Also your idea, which was mentioned in the link works fine, as the order of $3$ modulo $7$ is $6$. Another way (if you're familair with Galois Theory) is to notice that the maps defined by $a \to a^{3^i}$ for $i=0,1,\cdots 5$ are distinct automorphisms of $\mathbb{F}_3(a)$. This would mean that the Galois group of the extension $\mathbb{F}_3 \subset \mathbb{F}_3(a)$ has order at least $6$ and thus the extension degree is at least $6$. On the other side the extension has degree at most $6$, as $a$ is a root of $x^6+x^5+x^4+x^3+x^2+x+1$. This can help you conclude that the extension degree is $6$ and the field has $3^6$ elements.
A: (1) The question is unclear, the point being maybe the fact that there are some multiple roots, and the question wants then without multiplicity. The question is answered if we write $x^{21}-1 = (x^7-1)^3$, and the last polynomial $f=x^7-1$ has distinct roots. (The question may be seen as a help, as a hint.)
(2) The root $1$ already lives in the prime field. There is nothing to be done. Let $a$ be a root of $g=x^6+x^5+x^4+x^3+x^2+x+1$. This polynomial is irreducible. 
sage: F = GF(3)
sage: R.<x> = PolynomialRing(F)
sage: f = x^21 - 1
sage: f.factor()
(x + 2)^3 * (x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)^3
sage: g = (x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)
sage: g.factor()
x^6 + x^5 + x^4 + x^3 + x^2 + x + 1

The above sage code confirms this. Else, $a$ would be a root of a polynomial of degree $k$ among $ 2,3,4,5$, so would satisfy $a^{3^k}=a$. We get a contradiction as in the OP.
(3) The third question is clear, a lives in $\Bbb F_{3^6}$, but in no subfield. And there is the following splitting of $g$ over this bigger field:
sage: K.<a> = GF(3^6, modulus=g)
sage: g.base_extend(K).factor()
(x + 2*a) * (x + 2*a^2) * (x + 2*a^3) * (x + 2*a^4) 
* (x + a^5 + a^4 + a^3 + a^2 + a + 1) * (x + 2*a^5)

In other words, as expected,
$$g=\prod_{1\le k\le 6}(x-a^k)\ .$$
An other argumentation would be given by the observation that $x^7-1$ divides $x^{3^6}-x$, the last polynomial being the product of all $(x-s)$, $s$ in $\Bbb F_{3^6}$. (Because $7$ divides $3^6-1=728$, or Fermat.)
A: Question 1.
The question is asking for the number of distinct roots.
Question 2.
In the case that $a=1$, $\mathbb{F}_3(a)$ is not the splitting field of $g$ over $\mathbb{F}_3$.
