how to check if a number is a perfect square python

1x = int(input())
2print(int(round(x ** (1. / 3))) ** 3 == x)

1def is_cube(n):
2    guess = n**(1.0/3.0)
3    iguess = int(guess)
4    if iguess * iguess * iguess == n:
5        print(True, "it's cubed root is", iguess)
6        return
7    iguess = iguess + 1
8    if iguess * iguess * iguess == n:
9        print(True, "it's cubed root is", iguess)
10        return
11    print(False, "it's cubed root is", guess)
12    
13is_cube(9) # No
14is_cube(27) # Yes

1import math
2
3# Taking the input from user
4number = int(input("Enter the Number"))
5
6root = math.sqrt(number)
7if int(root + 0.5) ** 2 == number:
8    print(number, "is a perfect square")
9else:
10    print(number, "is not a perfect square")

1print(int(int(n)**.5))