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 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.