You could use x ** (1. / 3) to compute the (floating-point) cube root of x.

The slight subtlety here is that this works differently for negative numbers in Python 2 and 3. The following code, however, handles that:

def is_perfect_cube(x):
    x = abs(x)
    return int(round(x ** (1. / 3))) ** 3 == x

print(is_perfect_cube(63))
print(is_perfect_cube(64))
print(is_perfect_cube(65))
print(is_perfect_cube(-63))
print(is_perfect_cube(-64))
print(is_perfect_cube(-65))
print(is_perfect_cube(2146689000)) # no other currently posted solution
                                   # handles this correctly

This takes the cube root of x, rounds it to the nearest integer, raises to the third power, and finally checks whether the result equals x.

The reason to take the absolute value is to make the code work correctly for negative numbers across Python versions (Python 2 and 3 treat raising negative numbers to fractional powers differently).

UPDATE: Since python 3.11 one can use math.cbrt for the most accurate results

Python >= 3.11:

>>> import math
>>> math.cbrt(8)
2.0

OLD ANSWER:

You can use the power operator ** with fractions like:

Python3:

>>> 8**(1/3)
2.0

Python2:

>>> 8**(1.0/3)
2.0

In Python, you may find floating cube root by:

>>> def get_cube_root(num):
...     return num ** (1. / 3)
... 
>>> get_cube_root(27)
3.0

In case you want more generic approach to find nth root, you may use below code which is based on Newton's method:

# NOTE: You may use this function only for integer numbers
# k is num, n is the 'n' in nth root
def iroot(k, n):
    u, s = n, n+1
    while u < s:
        s = u
        t = (k-1) * s + n // pow(s, k-1)
        u = t // k
    return s

Compute one and multiply with the cube roots of 1?

cuberoots_of_1 = 1, complex(-.5, .75**.5), complex(-.5, -.75**.5)

def cuberoot(z):
    cuberoot = complex(z)**(1/3)
    return [cuberoot * cr1 for cr1 in cuberoots_of_1]

Test:

>>> cuberoot(-2 + 3j)
[(1.1532283040274223+1.0106429470939737j),
 (-1.4518566183526649+0.49340353410400484j),
 (0.2986283143252425-1.5040464811979786j)]

>>> for z in cuberoot(-2 + 3j):
        print(z**3, abs(z**3 - (-2 + 3j)))

(-1.999999999999999+3j) 1.1102230246251565e-15
(-1.999999999999999+3j) 1.1102230246251565e-15
(-1.9999999999999982+2.9999999999999996j) 1.831026719408895e-15

Doing the same with yours is less accurate:

(-1.999999999999996+3.000000000000002j) 4.572178254219406e-15
(-1.9999999999999933+3.000000000000004j) 7.768388458966724e-15
(-1.9999999999999956+3.0000000000000013j) 4.636427468134552e-15