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

This is another approach using the math module.

import math
num = int(input('Enter a number: '))
root = int(input('Enter a root: '))
nth_root = math.pow(num, (1/root))
nth_root = round(nth_root, 10)
print('\nThe {} root of {} is {}.'.format(root, num, nth_root))
decimal, whole = math.modf(nth_root)
print('The decimal portion of this cube root is {}.'.format(decimal))
decimal == 0

Line 1: Import math module.
Line 2: Enter the number you would like to get the root of.
Line 3: Enter the nth root you are looking for.
Line 4: Use the power function.
Line 5: Rounded to 10 significant figures to account for floating point approximations.
Line 6: Print a preview of the nth root of the selected number.
Line 7: Use the modf function to get the fractional and integer parts.
Line 8: Print a preview of decimal part of the cube root value.
Line 9: Return True if the cube root is an integer. Return False if the cube root value contains fractional numbers.