Proving that $\left[ \mathbb{Q} \left( \sqrt[3]{4+\sqrt{5}} \right ) : \mathbb{Q} \right] = 6$ Let $\alpha = \sqrt[3]{4+\sqrt{5}}$. I would like to prove that $\left[ \mathbb{Q} \left( \alpha \right ) : \mathbb{Q} \right] = 6$. We have $\alpha^3 = 4 + \sqrt{5}$, and so $(\alpha^3 - 4)^2 = 5$, hence $\alpha$ is a root of the polynomial $f(x)=x^6 - 8 x^3 + 11$.
I tried to prove with various approaches that $f(x)$ is irreducible over $\mathbb{Q}$ without success, so I devised the following strategy.
Since $x^2 - 5$ is irreducible over $\mathbb{Q}$, we have $\left[ \mathbb{Q} \left( \sqrt{5} \right ) : \mathbb{Q} \right] = 2$. Now from $\alpha^3 = 4 + \sqrt{5}$ we get $\sqrt{5} \in \mathbb{Q} \left( \alpha \right )$, so that $\mathbb{Q} \left( \sqrt{5} \right )$ is a subfield of $\mathbb{Q} \left( \alpha \right )$, $\mathbb{Q} \left( \alpha \right )=\mathbb{Q}\left( \sqrt{5}\right) \left( \alpha \right)$, and we have
\begin{equation}
\left[ \mathbb{Q} \left( \alpha \right ) : \mathbb{Q} \right] = \left[ \mathbb{Q} \left( \sqrt{5} \right)\left( \alpha \right ) : \mathbb{Q} \left (\sqrt{5} \right) \right] \left[ \mathbb{Q} \left( \sqrt{5} \right ) : \mathbb{Q} \right] .
\end{equation}
Now $\alpha$ is a root of the polynomial $g(x) \in \mathbb{Q} \left (\sqrt{5} \right) [ x ]$ given by $g(x) = x^3 - 4 - \sqrt{5}$. So to prove our thesis it is enough to prove that this polynomial is irreducible in $\mathbb{Q} \left (\sqrt{5} \right) [ x ]$. Being $g(x)$ of third degree, if it were not irreducible, its factorization would have at least one linear factor, so that $g(x)$ would have some root in $\mathbb{Q} \left (\sqrt{5} \right)$. Hence our problem boils down to show that there are no integers $m_0, m_1, n$, with $n \neq 0$, such that
\begin{equation}
\left( \frac{m_0}{n} + \frac{m_1}{n} \sqrt{5}\right)^3 = 4 + \sqrt{5},
\end{equation}
which gives
\begin{equation}
m_0^3 + 5 \sqrt{5} m_1^3 +3 \sqrt{5} m_0^2 m_1  + 15 m_0 m_1^2 = 4 n^3 + \sqrt{5} n^3,
\end{equation}
or
\begin{equation}
m_0^3 + 15 m_0 m_1^2 - 4 n^3 + \sqrt{5} \left( 5 m_1^3 +3 m_0^2 m_1  - n^3 \right)=0,
\end{equation}
which implies, being $\sqrt{5}$ irrational,
\begin{cases}
m_0^3 + 15 m_0 m_1^2 - 4 n^3 = 0, \\ 5 m_1^3 +3 m_0^2 m_1  - n^3 = 0.
\end{cases}
At this point I am stuck, because I do not know how to prove that this system admits the only integer solution $m_0 = m_1 = n = 0$.
Any help is welcome!

As $\Bbb Q(\sqrt 5)$ is a subfield, its automorphism $\sqrt 5\to-\sqrt 5$ can be extended, which means that $\sqrt[3]{4-\sqrt 5}\in\Bbb Q(\sqrt[3]{4+\sqrt 5})$ and also
$$ \sqrt[3]{11}=\sqrt[3]{4-\sqrt 5}\sqrt[3]{4+\sqrt 5}\in\Bbb Q(\sqrt[3]{4+\sqrt 5}).$$ Now, $[\Bbb Q(\sqrt[3]{11}):\Bbb Q]$ is clearly $3$. With a subfield of degree 2 and another of degree 3 over $\Bbb Q$, ouf field must be at least of degree $6$, hence exactly degree $6$.
