Let $F = \mathbb{Q}(\pi^3)$. Find a basis for $F(\pi)$ over $F$. Let $F = \mathbb{Q}(\pi^3)$. Find a basis for $F(\pi)$ over $F$.

How can I solve this. Can anyone help me please. Thanks.
 A: Hint: The number $\pi$ is transcendental. So this is the same problem as finding a basis for $F(x)$ over $F(x^3)$. 
A: Note that $\pi$ is algebraic over $\mathbb Q(\pi^3)$ with minimal polynomial $x^3-\pi^3$ (why?). Thus $\frac{\mathbb Q(\pi^3)[x]}{<x^3-\pi^3>}$ $\cong$ $\mathbb Q(\pi^3)(\pi)$ and $\{1,\pi,\pi^2\}$ is a basis for $\mathbb Q(\pi^3)(\pi)$ over $\mathbb Q(\pi^3)$.
The theorem behind this is as followed:
" Let $\mathbb F$ be a field  and let $p(x)\in\mathbb F[x]$ be irreducible over $\mathbb F$. If $a$ is a zero of $p(x)$ in some extension $\mathbb E$ of $\mathbb F$, then $\mathbb F(a)$ is isomorphic to $\frac{\mathbb F[x]}{<p(x)>}$. Furthermore, if $deg\{p(x)\}=n$, then every member of $\mathbb F(a)$ can be uniquely expressed as $$c_{n-1}a^{n-1}+c_{n-2}a^{n-2}+......c_{1}a+c_{0},$$
where $c_{0},c_{1},....,c_{n-1}\in\mathbb F$ "
So you can say that if $a$ is a zero of an irreducible polynomial over $\mathbb F$ of degree $n$, then the set $\{1,a,...,a^{n-1}\}$ is a basis for $\mathbb F(a)$ over $\mathbb F$.
A: Finding a basis for $F(x)$ over $F(x^3)$. 
This means you need to find a basis $v_1(x),\cdots,v_n(x)$ such that $p_1(x)v_1(x)+p_2(x)v_2(x)+\cdots+p_n(x)v_n(x)$ generates any polynomial of degree $n\in\mathcal{N}$ in $F(x)$. But $p_i(x)\in F(x^3)$ for any $i$, then $p_i(x)=a_nx^{3n}+a_{n-1}x^{3n-3}+\cdots+a_0$ where $a_i\in F$. 
It is clear that $\{v_1(x),v_2(x),v_3(x)\}=\{1,x,x^2\}$ is the easiest answer. 
