Consider the polynomial $p(x)=3^{3^3+1}x-x^{3^5}$ over the rationals $\mathbb{Q}$.
The solutions of $p(x)$ are given by $r_{n}:=\zeta^n \cdot 3^{\frac{3^3+1}{3^5-1}},~0 \leq n \leq 3^5-2$, where $\zeta$ is a primitive $3^5-1$ root of $1$ in $\mathbb{C}$.
Let us denote $K:=\mathbb{Q}[r_{0},r_{1},r_{2}, \dots, r_{3^5-2}]$.
I want to show the field generated by $3^{\frac{2}{3^5-1}} \cdot \zeta$ over $\mathbb{Q}$ is same as $K$.
Edit:
To make the context clear, let us consider another polynomial $q(x)=3^{3^2+1}x-x^{3^5}$ over the rationals $\mathbb{Q}$. The solutions of $q(x)$ are given by $s_{n}:=\zeta^n \cdot 3^{\frac{3^2+1}{3^5-1}},~0 \leq n \leq 3^5-2$, where $\zeta$ be a primitive $3^5-1$ root of $1$ in $\mathbb{C}$.
Let us denote $L:=\mathbb{Q}[s_{0},s_{1},s_{2}, \dots, s_{3^5-2}]$.
My purpose is to show $K=L$ by showing that both $K$ and $L$ are generated by same element over $\mathbb{Q}$ and that element is $3^{\frac{2}{3^5-1}} \cdot \zeta$.
This is why I asked my previous question.
Is there other way to show $K=L$ ?