Can $\cos (2\pi/7)$ be written as $p+\sqrt{q}+\sqrt[3]{r}, p,q,r\in \mathbb{Q}$? Is it possible to find $p,q,r \in \mathbb{Q}$ such that 
$$\cos \frac{2\pi}{7}=p+\sqrt{q}+\sqrt[3]{r}.$$
Assume we can find such $p,q,r$, then $\mathbb{Q}(\cos \frac{2\pi}{7}) \subseteq \mathbb{Q}(\sqrt{q},\sqrt[3]{r})$. I can show that $[\mathbb{Q}(\sqrt{q},\sqrt[3]{r}): \mathbb{Q}]=6$.(WLOG assuming $q$ and $r$ are square free and cube free resp.) I also know that $\mathbb{Q}(\cos \frac{2\pi}{7})/\mathbb{Q}$ is a cyclic extension (because $2\cos \frac{2\pi}{7}=\zeta_7+\zeta_7^{-1}$ where $\zeta_7$ is the primitive 7th root of unity and $\mathbb{Q}(\zeta_7)/\mathbb{Q}$ is a cyclic extension). But then I don't know how I can proceed, please help. 
 A: Suppose $\cos(2 \pi/7)=\sqrt[3] r+\sqrt q+p$ for some rational numbers $r,p,q.$
There are three cases:
First, suppose $q$ is not a square  and $r$ is not a cube in $\Bbb Q$. Then $\sqrt q+\sqrt[3]r$ has degree at least $6$ over $\Bbb Q$. This is because if $K$ is the Galois closure of $\Bbb Q(\sqrt q,\sqrt[3]r)$, then each embedding $\iota$ of $\Bbb Q(\sqrt q,\sqrt[3]r)$ into $K$ is determined by $\iota(\sqrt q)$ and $\iota(\sqrt[3] r)$, and $\iota(\sqrt q)$ and $\iota(\sqrt[3] r)$ can be any conjugate over $\Bbb Q$ of $\sqrt q$ and $\sqrt[3] r$ respectively. However, if $\iota_1$ and $\iota_2$ are distinct embeddings of $\Bbb Q(\sqrt q,\sqrt[3] r)$ into $K$, then it is clear from what I wrote above that $\iota_1(\sqrt q+\sqrt[3]r)\neq \iota_2(\sqrt q+\sqrt[3]r)$, and thus $\sqrt q+\sqrt[3]r$ has at least six conjugates over $\Bbb Q$. 
Therefore we cannot have $\cos(2\pi/7)=\sqrt q+\sqrt[3]r$ as $\cos(2\pi/7)$ has degree $3$ over $\Bbb Q$ since $$\cos(2\pi/7)=\frac{\zeta_7+\zeta_7^{-1}}{2},$$
and it is easy to check that $\zeta_7+\zeta_7^{-1}$ has three conjugates over $\Bbb Q$. Contradiction.
If $q$ is an square and $r$ is not a cube, then we have $\cos(2\pi/7)=\sqrt[3]r +b$ for some $b\in\Bbb Q$. The conjugates of $\cos(2\pi/7)$ over $\Bbb Q$ are all real, but the conjugates of $\sqrt[3] r+b$ over $\Bbb Q$ are $\sqrt[3] re^{2 \pi i/3}+b$ and $\sqrt[3] re^{4 \pi i/3}+b$, and this two complex numbers are not real. Contradiction.
Finally, if $r$ is a cube in $\Bbb Q$, we get a contradiction from the fact that $\cos(2\pi/7)$ has degree $3$ over $\Bbb Q.$
