Knowing that $\pi \in \mathbb{R}$ is a transcendent element over $\mathbb{Q}$. Find a subfield $\mathbb{F}$ of $\mathbb{R}$ such that $\pi$ is algebraic over $\mathbb{F}$, with $\deg(\pi, \mathbb{F}) = 3$.
I know the answer is $\mathbb{Q}(\pi^3)$ but I don't know if the path I took to get to the answer is correct.
If $f$ is algebraic over $\mathbb{F}$ then $f(\alpha) = 0$ and $$ x = \pi \implies x^3 = \pi^3 \implies x^3 - \pi^3 = 0 \implies \pi^3 = \alpha = x. $$
That's what I could think of, but I don't know if the argument is valid?
Grateful for the attention.
You seem to have the right idea, but your argument isn't so clear.
Let $\mathbb{F} = \mathbb{Q}(\pi^3)$, and define the polynomial $f \in \mathbb{F}[x]$ by $f(x) = x^3 - \pi^3$. Clearly, $\pi$ is a root of $f$. It's less clear (but still true!) that $\pi$ is not a root of a polynomial of lower degree over $\mathbb{F}$, so $\deg_{\mathbb{F}}(\pi) = 3$.