Is cube root integer?

Question

This seems to be simple but I cannot find a way to do it. I need to show whether the cube root of an integer is integer or not. I used is_integer() float method

Answer 1

For small numbers (<~10¹³ or so), you can use the following approach:

def is_perfect_cube(n):
    c = int(n**(1/3.))
    return (c**3 == n) or ((c+1)**3 == n)

This truncates the floating-point cuberoot, then tests the two nearest integers.

For larger numbers, one way to do it is to do a binary search for the true cube root using integers only to preserve precision:

def find_cube_root(n):
    lo = 0
    hi = n
    while lo < hi:
        mid = (lo+hi)//2
        if mid**3 < n:
            lo = mid+1
        else:
            hi = mid
    return lo

def is_perfect_cube(n):
    return find_cube_root(n)**3 == n