Show that $\pi \notin Q(\pi^3)$ 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.
 A: The following general result might interest you.
If $k$ is a field and if $\phi(x)=\frac {f(x)}{g(x)}\in k(x)$ is a rational function with $f(x),g(x)\in k[x]$ relatively prime polynomials (not both constant), then the extension of fields $k(\phi (x))\subset k(x)$ has degree $$[k(x):k(\phi (x))]=\text {max}\:(\text {deg} \; f(x),\text {deg} \; g(x))$$   This of course implies (if you know that $\pi$ is transcendental and thus may play the role of the indeterminate $x$) that $[\mathbb Q(\pi):\mathbb Q(\pi^3]=3$ and thus a fortiori that $\pi \notin \mathbb Q(\pi^3)$ .
Bibliography
The displayed formula can be found in  Theorem 8.36, page 614 of Jacobson's Basic Algebra II. 
A: 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 $[L\colon K] = 3$.
A: From $$\pi = \frac{a_n\pi^{3n}+\ldots+a_1\pi^3+a_0}{b_m\pi^{3m}+\ldots +b_1\pi^3+b_0}$$
with $a_n\ne0$, $b_m\ne0$
we obtain a polynomial equation for $\pi$:
$$\tag1(b_m\pi^{3m+1}+\ldots +b_1\pi^4+b_0\pi)-(a_n\pi^{3n}+\ldots+a_1\pi^3+a_0)=0.$$
If $n>m$, this is of degree $3n$ with leading coefficient $-a_n\ne 0$, if $n\le m$ this is of degree $3m+1$ with leading coefficient $b_m\ne0$.
Hence $(1)$ shows that $\pi$ is algabraic, which it isn't.

From
$$\sqrt 2=\frac{a_n\pi^{n}+\ldots+a_1\pi+a_0}{b_m\pi^{m}+\ldots +b_1\pi+b_0},$$
we obtain
$$2=\frac{a_n^2\pi^{2n}+\ldots+2a_0a_1\pi+a_0^2}{b_m^2\pi^{2m}+\ldots +2b_0b_1\pi+b_0^2},$$
hence 
$$\tag2(a_n^2\pi^{2n}+\ldots+a_0^2)-2(b_m^2\pi^{2m}+\ldots +b_0^2)=0.$$
Since $\pi$ is transcendental, this must be the zero polynomial, i.e. everything cancels.
Especially, we must have $n=m$ and $a_n^2-2b_m^2=0$.
But then $\sqrt 2=\left\vert\frac{a_n}{b_m}\right\vert\in\mathbb Q$.
