<s>[INST] 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 $2-\sqrt{5}$.
 [/INST] A calculation shows that $(1+\sqrt{5})^3=16+8\sqrt{5}$, so the real cube root of $2+\sqrt{5}$ is $\frac{1}{2}(1+\sqrt{5})$.
Similarly, or by using conjugation, the real cube root of $2-\sqrt{5}$ is $\frac{1}{2}(1-\sqrt{5})$.
Add. 
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