TITLE: Show that $\pi \notin Q(\pi^3)$
QUESTION [9 upvotes]: As the title says. I think a proof by contradiction is the most natural thing. Suppose $\pi \in Q(\pi^3)$. Then
\begin{equation}
\pi = \frac{a_n(\pi^3)^n+\cdots+a_1\pi^3+a_0}{b_m(\pi^3)^m+\cdots+b_1\pi^3+b_0} \text{.}
\end{equation}
Not sure how to proceed from here though.
I also have a related question, that is to show $\sqrt{2} \notin Q(\pi)$. I think if I can solve either one of them, I can solve the other. Any help would be much appreciated.

REPLY [3 votes]: I will prove the problem as a simple case of what Georges describes.
Let $k$ be a field.
Let $L = k(x)$ be the rational function field with one variable $x$(we may take $x = \pi$ when $k = \mathbb{Q}$).
Let $K = k(x^3)$.
Then $L = K(x)$.
Let $X^3 - a \in K[X]$, where $a = x^3$.
Clearly $X^3 - a$ cannot have a root in $K$ considering the degree of a root if any.
Hence it is irreducible over $K$.
Hence $[L\colon K] = 3$.