Minimal polynomial: is $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1$? I was wondering about the minimal polynomial of real number
$$u=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$$
over field $\mathbb{Q}$.
As you can see here, I worked out that $u$ is a root of monic rational polynomial $x^3+3x-4$. This is not irreducible:
$$x^3+3x-4=\left(x-1\right)\left(x^2+x+4\right)$$
and the second, quadratic, factor has complex roots $\frac{-1\pm i\sqrt{15}}{2}$.
Can I claim that the minimal polynomial of $u$ over $\mathbb{Q}$ is $x-1$?
In other words: does this prove that $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1$?
In case it does, how can we demonstrate $u=1$ working on the roots in a way different from how I did in the link?
Edit:
$\sqrt[3]{2-\sqrt{5}}$ is meant to be the real cube root of $2-\sqrt{5}$.

Here's another way of looking at this through a reverse lens:
Let's solve $u^3+3u-4=0$ by Cardano's method, putting $u=x+y$.
Then $(x+y)^3-3xy(x+y)-(x^3+y^3)=0$ and we require:
$$x^3+y^3=4$$ and $$-3xy = 3 \text { so that }xy=x^3y^3=-1$$
Then we note that $x^3$ and $y^3$ are roots of the quadratic $$z^2-4z-1=0$$So that $$z=\frac{4\pm\sqrt{16+4}}{2}=2\pm \sqrt 5$$ 
Whence $$u=\sqrt[3]{2+\sqrt 5}+\sqrt[3]{2-\sqrt 5}$$ where the cube roots have to be compatible with the constraint $xy=-1$. There are three possibilities and three roots of the cubic. Since $u=1$ is a root of the cubic, one choice will give the root $u=1$.
Since the cubic has only one real root, choosing the real cube roots will give $u=1$.
