# Cube root of a number is equal to its own cube. What are all the possibilities for this number?

Apr 20, 2016

At least $0 , - 1$ and $1$.

With $A r g \left(z\right) \in \left(- \pi , \pi\right]$ there are additional Complex solutions:

$- \frac{\sqrt{2}}{2} \pm \frac{\sqrt{2}}{2} i$

#### Explanation:

Given:

${x}^{3} = \sqrt{x}$

Cubing both sides we get:

${x}^{9} = x$

Note that cubing both sides may (and actually does) introduce spurious solutions.

One solution is $x = 0$, which is a solution of the original equation.

If $x \ne 0$ then we can divide both sides by $x$ to get:

${x}^{8} = 1$

This has Real solutions $x = \pm 1$, both of which are solutions of the original problem, if we follow the normal convention that $\sqrt{1} = 1$ and $\sqrt{- 1} = - 1$.

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Any possible Complex solutions of this octic equation can be expressed in the form:

$\cos \left(\frac{k \pi}{4}\right) + i \sin \left(\frac{k \pi}{4}\right)$ for $k = 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7$

That is:

$x = \left\{\begin{matrix}1 \\ \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i \\ i \\ - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i \\ - 1 \\ - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i \\ - i \\ \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i\end{matrix}\right.$

The question as to whether any of these is a solution of the original equation depends on what you mean by $\sqrt{x}$ if $x \in \mathbb{C}$.

Suppose, consistent with $A r g \left(z\right) \in \left(- \pi , \pi\right]$, we define:

$\sqrt{\cos \theta + i \sin \theta} = \cos \left(\frac{\theta}{3}\right) + i \sin \left(\frac{\theta}{3}\right)$

when $- \pi < \theta < \pi$

Most of the solutions of ${x}^{8} = 1$ are not solutions of ${x}^{3} = \sqrt{x}$, but the following two are:

$x = \left\{\begin{matrix}- \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i \\ - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i\end{matrix}\right.$

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Note that if you prefer $A r g \left(z\right) \in \left[0 , 2 \pi\right)$ then the valid solutions apart from $0 , 1 \mathmr{and} - 1$ are:

$x = \left\{\begin{matrix}- \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i \\ - i\end{matrix}\right.$

Apr 20, 2016

is there a more systematic way?

#### Explanation:

${x}^{8} = {e}^{2 n i \pi}$
$x = {e}^{n i \frac{\pi}{4}}$ where $n = 1 , 8$

Apr 20, 2016

The solutions are $\pm 1. \pm i , \frac{\pm 1 \pm i}{\sqrt{2}}$

#### Explanation:

Let the number be x.

${x}^{\frac{1}{3}} = {x}^{3}$.
${x}^{3 - \frac{1}{3}} = 1$
${x}^{\frac{8}{3}} = 1$
$x = {1}^{\frac{3}{8}} . = {\left({1}^{3}\right)}^{\frac{1}{8}} = {1}^{\frac{1}{8}}$

$S o , x = {\left({e}^{i 2 n \pi}\right)}^{\frac{1}{8}} = {e}^{\frac{i 2 n \pi}{8}}$, n = any integer, including 0.

The values for n = 0, 1, 2.., 7 are repeated, in a cycle, for other values of n.

The eight distinct values are determinate and are listed in the answer.

Importantly, observe that x = 0 is not a solution. .