Find the least $n>1$ such that $\cos(\pi/n)$ can NOT be written as $a+\sqrt{b} + \sqrt[3]{c}$ for $a,b,c\in \mathbb{Q}$ I'm currently learning algebraic number theory, and this is an exercise of the first chapter of my lecture (regarding symmetric polynomials, minimal polynomials, conjugates) that I've been struggling with for the past few days.
Uptill now, I've been completely stuck: $a+\sqrt{b}+\sqrt[3]{c}$ should (but I'm not sure how to rigorously prove that) have a minimal polynomial of degree at most $6$, so I think I should look for $n$ with $\cos(\pi/n)$ of degree $n$, but how to move forward?
Thank you.
 A: I think I found the solution, thanks to @user994373.
We want to exhib the least $n$ such that $\cos(\pi/n)$ can not be written as $a+\sqrt{b} + \sqrt[3]{c}$ for some rationals $a,b,c$.
$\textbf{First step : $\cos(\pi/n)$ has only real conjugates}$.
We define Chebychev polynomials of second kind as the polynomial verifying :
$$\left\{
\begin{array}{ll}
U_0 = 1\\
U_1 = 2X\\
U_{n+1} = 2XU_n - U_{n-1}
\end{array}\right.$$
By induction, these polynomials have only rational coefficients, and $\deg U_n = n$.
Moreover, one can prove that $$\forall \theta\in \mathbb{R}\backslash \pi\mathbb{Z},\quad U_{n-1}(\cos(\theta)) = \frac{\sin(n\theta)}{\sin(\theta)}$$
It follows that $\cos(\pi/n),\cos(2\pi/n),\cos(3\pi/n),\dots,\cos((n-1)\pi/n)$ are the $n-1$ roots of $U_{n-1}$. Thus, $\cos(\pi/n)$ only has real conjugates.
$\textbf{Second step : conditions for $a+\sqrt{b}+\sqrt[3]{c}$ to only have real conjugates}$.
As a corollary of the fundamental theorem of symmetric polynomials, the conjugates of $a+\sqrt{b}+\sqrt[3]{c}$ lie in $$ S = \{a\pm \sqrt{b}+\sqrt[3]{c},\; a\pm \sqrt{b}+j\sqrt[3]{c},\; a\pm \sqrt{b}+j^2\sqrt[3]{c}\}$$
where $j:= \exp(2i\pi/3)$.
If $b=0$, it is enough for $\sqrt[3]{c}$ to not be rational so that $a+\sqrt[3]{c}$ admits at least two conjugates (both in $S$), one at least being non-real.
If $b>0$, we see that $$\begin{align*}P(X) &= \big(X - (a+\sqrt{b}+\sqrt[3]{c})\big)\big(X - (a-\sqrt{b}+\sqrt[3]{c})\big)\\ &= X^2 - 2(a+\sqrt[3]{c})X + K\end{align*}$$
is not in $\mathbb{Q}[X]$ if $\sqrt[3]{c}\not\in \mathbb{Q}$. Therefore, $a+\sqrt{b}+\sqrt[3]{c}$ has at least $3$ conjugates (all in $S$) if $\sqrt[3]{c}\not\in \mathbb{Q}$, and at least one of them is then non-real.
As a result, the condition for $a+\sqrt{b}+\sqrt[3]{c}$ to having only real conjugates is that $\sqrt[3]{c}\in \mathbb{Q}$ (which is equivalent to saying $c=0$, without loss of generality, as $\sqrt[3]{c}$ can be "absorbed" in $a$).
$\textbf{Third step : conclusion.}$
Because $\cos(\pi/n)$ only has real conjugates, if it can be expressed as $a+\sqrt{b}+\sqrt[3]{c}$, then $c=0$. But the minimal polynomial of $a+\sqrt{b}$ is at most of degree $2$. Similarly, if the minimal polynomial of $\cos(\pi/n)$ is of degree $1$ or $2$, then $\cos(\pi/n)$ can be expressed as $a+\sqrt{b}$ for some rationals $a$ and $b$.
Therefore, we're looking for the least $n$ such that $\cos(\pi/n)$ has a minimal polynomial of degree $3$ (or more, if none have degree $3$).
With a bit of time, we find that $n=7$ works (the minimal polynomial of $\cos(\pi/7)$ is $8 x^3 - 4 x^2 - 4 x + 1$).
