How to find cube root using Python? [duplicate]

Here is how, this is the best way, I have found:

x = int(raw_input("Enter an integer: ")) for ans in range(0, abs(x) + 1): if ans ** 3 == abs(x): break if ans ** 3 != abs(x): print x, 'is not a perfect cube!' else: if x < 0: ans = -ans print 'Cube root of ' + str(x) + ' is ' + str(ans) 

Is there a better way, preferably one that avoids having to iterate over candidate values?

3 Answers

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).

The best way is to use simple math

>>> a = 8 >>> a**(1./3.) 2.0 

EDIT

For Negative numbers

>>> a = -8 >>> -(-a)**(1./3.) -2.0 

Complete Program for all the requirements as specified

x = int(input("Enter an integer: ")) if x>0: ans = x**(1./3.) if ans ** 3 != abs(x): print x, 'is not a perfect cube!' else: ans = -((-x)**(1./3.)) if ans ** 3 != -abs(x): print x, 'is not a perfect cube!' print 'Cube root of ' + str(x) + ' is ' + str(ans) 

def cube(x): if 0<=x: return x**(1./3.) return -(-x)**(1./3.) print (cube(8)) print (cube(-8)) 

Here is the full answer for both negative and positive numbers.

>>> 2.0 -2.0 >>> 

Or here is a one-liner;

root_cube = lambda x: x**(1./3.) if 0<=x else -(-x)**(1./3.)