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

##### 3 Answers

At least

With

#-sqrt(2)/2+-sqrt(2)/2i#

#### Explanation:

Given:

#x^3=root(3)(x)#

Cubing both sides we get:

#x^9 = x#

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

One solution is

If

#x^8 = 1#

This has Real solutions

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

#cos((k pi)/4) + i sin((k pi)/4)# for#k = 0, 1, 2, 3, 4, 5, 6, 7#

That is:

#x = { (1), (sqrt(2)/2+sqrt(2)/2i), (i), (-sqrt(2)/2+sqrt(2)/2i), (-1), (-sqrt(2)/2-sqrt(2)/2i), (-i), (sqrt(2)/2-sqrt(2)/2i) :}#

The question as to whether any of these is a solution of the original equation depends on what you mean by

Suppose, consistent with

#root(3)(cos theta + i sin theta) = cos (theta/3) + i sin (theta/3)#

when

Most of the solutions of

#x = { (-sqrt(2)/2+sqrt(2)/2i), (-sqrt(2)/2-sqrt(2)/2i) :}#

Note that if you prefer

#x = { (-sqrt(2)/2+sqrt(2)/2i), (-i) :}#

is there a more systematic way?

#### Explanation:

The solutions are

#### Explanation:

Let the number be x.

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. .