Question

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

Answer 1

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

With `Arg(z) in (-pi, pi]` there are additional Complex solutions:

`-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 `x=0`, which is a solution of the original equation.

If `x != 0` then we can divide both sides by `x` to get:

`x^8 = 1`

This has Real solutions `x = +-1`, both of which are solutions of the original problem, if we follow the normal convention that `root(3)(1) = 1` and `root(3)(-1) = -1`.

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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 `root(3)(x)` if `x in CC`.

Suppose, consistent with `Arg(z) in (-pi, pi]`, we define:

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

when `-pi < theta < pi`

Most of the solutions of `x^8=1` are not solutions of `x^3 = root(3)(x)`, but the following two are:

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

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Note that if you prefer `Arg(z) in [0, 2pi)` then the valid solutions apart from `0, 1 and -1` are:

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

Answer 2

is there a more systematic way?

Explanation:

`x^8 = e^(2nipi)`
`x = e^(ni pi/4)` where `n=1,8`

Answer 3

The solutions are `+-1. +-i, (+-1+-i)/sqrt2`

Explanation:

Let the number be x.

`x^(1/3)=x^3`.
`x^(3-1/3)=1`
`x^(8/3)=1`
`x=1^(3/8). =(1^3)^(1/8)=1^(1/8)`

`So, x=(e^(i2n pi))^(1/8)=e^((i2npi)/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. .