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.
Hint: The number $\pi$ is transcendental. So this is the same problem as finding a basis for $F(x)$ over $F(x^3)$.
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.