posted before $n=7$.

Rust wrote: posted before $n=7$. Can you give a link?

A topic with a bad title.

The degree of the minimal polynomial of $\cos{\pi/7}$ is $3$. To those with more experience, it is obvious that $p+\sqrt{q}+\sqrt[3]{r}$ cannot satisfy an integer cubic polynomial unless $\sqrt{q}\in \mathbb{Q}$, i.e. $q$ is a square. For the others, once you have observed and showed this rather standard fact, everything follows smoothly. For then remains to show that $\cos{\pi/7}$ does not take the form $p+\sqrt[3]{r}$. The latter number has algebraic conjugates $p+\omega^{j}\sqrt[3]{r}$ for $j=0,1,2$, $\omega = e^{2\pi i/3}$, which gives you the minimal polynomial for $p+\sqrt[3]{r}$; and writing down the minimal polynomial of $\cos{\pi/7}$, you'll be done by comparing the coefficients. To show the well-known fact that $p+\sqrt{q}+\sqrt[3]{r}$ can't satisfy an integer cubic unless $q$ is a square, one way is to look at this thread https://artofproblemsolving.com/community/c6h27089. The "theorem" I had stated there implies this a very, very special case .

Does there exist a more elementary solution?