root of $x^3+x+1$ over $\mathbb{F}_5$ What is the simplest radical expression of some root $a \in \overline{\mathbb{F}_5}$ of the polynomial $x^3+x+1  \in \mathbb{F}_5[x]$? I wonder if one can simplify the general formulas in this special case. I think $\sqrt[3]{\sqrt{2} - 3} - \sqrt[3]{\sqrt{2} + 3}$ is ok for the beginning, but this minus in the middle and the nested roots make it rather complicated.
 A: $\mathbb F_{5^3}$ is a cubic extension of $\mathbb F_5$, and the first thing to ask if it we can write it as $\mathbb F_5(\sqrt[3]a)$. Unfortunately, 5 is not congruent to 1 mod 3, so $x \mapsto x^3$ is a permutation of $F_5$, and the polynomials $X^3 - a$ are always reducible. 
In order to express $\alpha$ with radicals, you will need a cube root, so we have no choice but to go to $F_{5^2}$ first by taking a square root, then use a cube root to go to $F_{5^6}$, and then express $\alpha$ just like you did.
Now, rearranging your expression, we have $\alpha = \sqrt[3]{2 + \sqrt{2}} + \sqrt[3]{2 - \sqrt{2}}$. It is clear it is in $\mathbb F_{5^3}$ because it is invariant by the automorphism of $\mathbb F_{5^2}$ switching $\sqrt 2$ with $ - \sqrt 2$.
However it is unclear because we pick two different cube roots and someone might make a mistake and not take conjugated roots. It may be better to write $\sqrt[3]{2 - \sqrt{2}}$ = $\frac 1 {\sqrt[3]{2 + \sqrt{2}}}$ instead, so $\alpha = \sqrt[3]{2 + \sqrt{2}} + \frac 1 {\sqrt[3]{2 + \sqrt{2}}}$
A: I think, this cannot be much simplified.
All elements ($0,\pm 1,\pm 2$) are cubic roots, so one $\sqrt[3]\ $ is not enough. The simplest you can get then (if you insist on root signs) is to introduce a $\sqrt\ $.
But our field extension $\mathbb F_{125} \cong \mathbb F_5(a)$ has degree $3$, so doesn't contain any pure square root, so one additionally need a cubic root finally, and this minus trick with the conjugate, to get back of degree $3$..
