Getting all solutions of (-1)^(2/3) ?

Questions
,
1

How do I find all solutions of (-1)^(2/3)?

If I evaluate

In[26]:= N[(-1)^(2/3)]
Out[26]= -0.5 + 0.866025 I

Or if I try to solve

In[21]:= Solve[x^(3/2) == -1, x, Complexes] // N
Out[21]= {{x -> -0.5 + 0.866025 I}}

How do I get all the complex numbers that are solutions?

2

Not sure why Solve fails but Reduce seems to work:

Solve[x^(3/2) == -1, x, Method -> Reduce]

(*  {{x -> -(-1)^(1/3)}, {x -> (-1)^(2/3)}}  *)

(As you may know, x^(3/2) is the principal root, and so we don’t get all solutions to the rationalized equation, (x^3) == (-1)^2.)

3

A variation:

Solve[{y^2 == x, y^3 == -1}, {x, y}]
% // N
4

Thank you very much!
So I need to specify it like this:

In[1]:= Solve[x^3 == (-1)^2, x]
Out[1]= {{x -> 1}, {x -> -(-1)^(1/3)}, {x -> (-1)^(2/3)}}
5

It really depends on what you mean by (-1)^(2/3). It is not obvious.