Why are the roots of $x^3+2=0 $: $-2^{1/3}, (-2)^{1/3}$, and $-(-1)^{2/3} 2^{1/3}$?

Question

For some reason I can't understand why the roots of $x^3+2=0$ are:

$$x = -\sqrt[3]{2}$$ $$x = \sqrt[3]{-2}$$ $$x = -(-1)^\frac{2}{3} \sqrt[3]{2}$$

I thought it was easier to solve this equation by doing this (I'm probably wrong):

$$x^3+2=0 \iff x^3=-2 \iff x= \pm \sqrt[3]{-2} $$

I put the function on a graph and saw that $x = -\sqrt[3]{2}$ was indeed the real root of it. However, I can't understand why.

(I started learning imaginary numbers and complex roots a few days ago so that might explain my reasoning.)

Thank you!

Kenny

Answer 1

They are roots because if you plug them into the equation, they give you zero (check it!). If you are asking HOW to find roots of a polynomial of this form, I would suggest factoring out $(x+\sqrt[3]{2})$ first (because you know that $- \sqrt[3]{2}$ is a root) and then using the quadratic formula to find the other roots.

Answer 2

Note the three solutions of $x^3+2=0$ are \begin{align*} &\{-\sqrt[3]{2},+\sqrt[3]{-1}\cdot\sqrt[3]{2},-\sqrt[3]{-1}\cdot\sqrt[3]{2}\}=\left\{\left.-\sqrt[3]{2}\cdot \exp\left(\frac{2k\pi i}{3}\right)\right|k=0,1,2\right\} \end{align*} with $\exp\left(\frac{2k\pi i}{3}\right), k=0,1,2$ the three roots of unity.

• The key to the three different solutions are the three roots of unity, the vertices of a regular $3$-gon at $\left\{1,\pm \exp\left(\frac{2\pi i}{3}\right)\right\}$ scaled by $\sqrt[3]{2}$ and rotated by $\pi$.

\begin{align*} z^3=1\quad\rightarrow\quad\left(\frac{z}{\sqrt[3]{2}}\right)^3=1\quad\rightarrow\quad\left(-\frac{z}{\sqrt[3]{2}}\right)^3=1\\ \end{align*}

which is

\begin{align*} z^3=1\quad\rightarrow\quad z^3=2\quad\rightarrow\quad z^3=-2\\ \end{align*}

• In general we consider $x^n=1$ and $n$ roots of unity forming a regular $n$-gon.

• The special case with only two solutions ($\pm$) is valid for quadratic equations as in \begin{align*} x^2+2=0\quad\iff \quad x^2=-2 \quad\iff \quad x= \pm \sqrt{-2} \end{align*}