Find all non-real values of $x$ if $x^3+1/x^3=110$

Question

$$x^3+1/x^3=110$$ $$x^6-110x^3+1=0$$ $$x^3=55\pm 12\sqrt{21}$$ $$x=\omega^t\sqrt[3]{55\pm 12\sqrt{21}},\ t=1\text{ or }2$$ 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.

Answer 1

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.

Answer 2

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 $$ \root3\of{55+12\sqrt{21}}=\frac{5+\sqrt{21}}2. $$

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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).