Find all non-real values of $x$ if $x^3+1/x^3=110$ $$x^3+1/x^3=110$$
$$x^6-110x^3+1=0$$
$$x^3=55\pm 12\sqrt{21}$$
$$x=\omega^t\sqrt[3]{55\pm 12\sqrt{21}},\ t=1\text{ or }2$$
Here $\omega$ is a primitive third root of unity. This looks a little unwieldy – how can I simplify this?
Looking for a (pre-calculus) level solution.
 A: This uses a few basic facts of algebraic number theory, so may not be the most accessible route. It does lead to an answer :-)
You need to play with the units of the ring $O_K$ of algebraic integers of the field $K=\Bbb{Q}(\sqrt{21})$. It is a basic fact that 
$$
O_K=\Bbb{Z}[\frac{1+\sqrt{21}}2].
$$
The units of that ring form a group $O_K^*=C_2\times\langle u\rangle$, where $C_2$
is the cyclic group of order two generated by $-1$, and $u$ is a so called fundamental unit. 
The number $55+12\sqrt{21}$ is an element of that group, and its cube root can be denested if and only if it is a cube of an element of that group.
Unless I made a mistake the fundamental unit of that group is $u=(5+\sqrt{21})/2$. And today is our lucky day, because
$$
u^3=55+12\sqrt{21}.
$$
So the answer is
$$
\root3\of{55+12\sqrt{21}}=\frac{5+\sqrt{21}}2.
$$

You don't necessarily need to learn that much algebraic number theory to get to the finish line. Familiarity with Pell equations, here $x^2-21y^2=1$ (and $x^2-21y^2=4$), will do.
See


*

*this wikipedia article for basics about fundamental units,

*this for basics of Pell equations, and

*this for continued fractions (a tool for finding the fundamental unit of quadratic number field - in this case trial and error was faster).

A: A pre-calculus route is the following. Let $v=x+1/x$. Then
$$
x^3+\frac1{x^3}=(x+\frac1x)^3-3(x+\frac1x)=v^3-3v.
$$
So we have the equation
$$
v^3-3v=110.
$$
This has $v=5$ as an obvious solution (rational root test, or just observation, or, if everything else fails, full Cardano).
But the solutions of
$$
x+\frac1x=5
$$
are
$$
x=\frac{5\pm\sqrt{21}}2.
$$
With the aid of this you can then easily answer the original question. After all, these solutions must be equal to the real solutions $x=\root3\of{55\pm12\sqrt{21}}$ that you found.
