How to show this is the minimal polynomial I'm trying to the following problem. But I can't show some irreducibility of the polynomials.
Put $L=\mathbb{C}(X,Y,Z)$, $\omega=\frac{-1+\sqrt{-3}}{2}$.
Define two automorphism $\sigma, \tau$ of $L$ over $\mathbb{C}$ by
$\sigma(X)=Y, \sigma(Y)=Z, \sigma(Z)=X, \tau(X) = X,\tau(Y)=\omega Y,\tau(Z)=\omega^2 Z$.
$K :=\{x\in L|\sigma(x)=x, \tau(x)=x\}$. Then,
(1) Find the minimal polonominal of $X$ over $K$. 
(2) Calculate $[K(X):K]$.
(3) Calculate $[L:K]$.
First of all, by symmetry of $\sigma$, I suppose the minimal polynomial is something symmetric. So I calculate
$(T^3-X^3)(T^3-Y^3)(T^3-Z^3)=T^9-(X^3+Y^3+Z^3)T^6+(X^3Y^3+Y^3Z^3+Z^3X^3)T^3-X^3Y^3Z^3 =:p(T)$
Since $X^3+Y^3+Z^3,X^3Y^3+Y^3Z^3+Z^3X^3,X^3Y^3Z^3\in K$, this polynomial $p(T)\in K[T]$ and satisfy $p(X)=0$. So I think this is the minimal polynomial. But I can't show this is irreducible. This is the first question.
Next, to calculate $[L:K]$, I remarked $K(X,Y)=L$ because $XYZ\in K$. Then, I think the minimal polynomial of $Y$ over $K(X)$ might be also $p(T)$. But I can't show that this polynomial is irreducible. This is the second question.
 A: You are on a good start. I would argue as follows. I write $X=u_1$, $Y=u_2$, $Z=u_3$ to make a few things notationally simpler.


*

*Show that the group $G$ generated by $\sigma$ and $\tau$ consists of the 27 automorphisms of the form $u_i\mapsto \omega^{a_i}u_{\alpha(i)}, i=1,2,3$, where $\alpha$ is a permutation in the subgroup $\langle(123)\rangle\le S_3$ (three choices), and where the vector of exponents $(a_1,a_2,a_3)$ is in the zero sum subgroup $P\le\Bbb{Z}_3^3$ (nine choices).

*In light of item 1 we see that $X=u_1$ has exactly nine conjugates (its orbit under $G$), namely $\omega^ju_i$, $i,j\in\{1,2,3\}$. Therefore its minimal polynomial must have degree nine. As we are in characteristic zero all those conjugates are simple zeros of the minimal polynomial. This leads to your minimal polynomial.

*A known result of field theory is that if $\alpha$ is algebraic over a field $K$ and its minimal polynomial has degree $n$, then $[K(\alpha):K]=n$. 

*A known result of Galois theory is that if $L$ is a field, and $G$ is a finite group of automorphisms of $L$, then $[L:K]=|G|$, where $K$ is the fixed field of $G$.

