Extension of $\mathbb{Q}$ over which $\pi$ is algebraic of degree 3?

Question

I think $\pi$ is algebraic of degree 3 over $\mathbb{Q}(\pi^3)$. To prove it, I need to show that $\pi \notin \mathbb{Q}(\pi^3)$ (which implies that $x^3-\pi^3$ is the minimum polynomial of $\pi$ over $\mathbb{Q}(\pi^3)$). I have no experience solving this kind of problem for transcendental elements like $\pi$.

Answer 1

Suppose $\pi\in\mathbb{Q}(\pi^3)$. That means that there are some polynomials $f(x)$ and $g(x)$ with rational coefficients (with $g(x)\neq 0$) such that $\pi=f(\pi^3)/g(\pi^3)$, or $\pi g(\pi^3)-f(\pi^3)=0$. But this is then a polynomial in $\pi$ which vanishes, so since $\pi$ is transcendental, it must be identically $0$. This is impossible, since only powers of $\pi$ divisible by $3$ appear in $f(\pi^3)$, and only powers of $\pi$ which are $1$ mod $3$ appear in $\pi g(\pi^3)$, so there can be no cancellation of terms.