Python Program to Find Cube Root of a Number

You can find the cube root of a number in Python using the expression number ** (1/3) or by using pow(number, 1/3).

Examples

Input27

Output3.0

Input64

Output4.0

Input-8

Output-2.0

How to Think About It

To find the cube root of a number, think about what number multiplied by itself three times equals the original number. We can use the power operator ** with the fraction 1/3 to calculate this directly in Python.

Algorithm

1

Get the input number from the user or as a variable.

2

Calculate the cube root by raising the number to the power of 1/3.

3

Return or print the result.

Code

python

number = float(input("Enter a number: "))
if number >= 0:
    cube_root = number ** (1/3)
else:
    cube_root = -(-number) ** (1/3)
print(f"Cube root of {number} is {cube_root}")

Output

Enter a number: 27
Cube root of 27.0 is 3.0

Dry Run

Let's trace the input 27 through the code

1

Input number

User enters 27, so number = 27.0

2

Calculate cube root

cube_root = 27.0 ** (1/3) = 3.0

3

Print result

Output: Cube root of 27.0 is 3.0

Step	Variable	Value
1	number	27.0
2	cube_root	3.0
3	output	Cube root of 27.0 is 3.0

Why This Works

Step 1: Using power operator

The cube root of a number is the same as raising it to the power of 1/3, so number ** (1/3) calculates this directly.

Step 2: Handling negative numbers

Raising a negative number to the power 1/3 returns a complex number in Python, so for negative inputs, we adjust the calculation to keep the result real.

Step 3: Printing the result

We use print with an f-string to show the original number and its cube root clearly.

Alternative Approaches

Using pow() function

python

number = float(input("Enter a number: "))
if number >= 0:
    cube_root = pow(number, 1/3)
else:
    cube_root = -pow(-number, 1/3)
print(f"Cube root of {number} is {cube_root}")

This is similar to the power operator but uses a built-in function; readability is a matter of preference.

Handling negative numbers explicitly

python

number = float(input("Enter a number: "))
if number >= 0:
    cube_root = number ** (1/3)
else:
    cube_root = -(-number) ** (1/3)
print(f"Cube root of {number} is {cube_root}")

This approach ensures the cube root of negative numbers is a real number, avoiding complex results.

Complexity: O(1) time, O(1) space

Time Complexity

Calculating the cube root using the power operator or pow() is a constant time operation, so it is O(1).

Space Complexity

The calculation uses a fixed amount of memory for variables, so space complexity is O(1).

Which Approach is Fastest?

Using the power operator ** or pow() are equally fast; handling negatives explicitly adds minimal overhead but improves correctness.

Approach	Time	Space	Best For
Power operator **	O(1)	O(1)	Simple positive numbers
pow() function	O(1)	O(1)	Equivalent to power operator, preference-based
Explicit negative handling	O(1)	O(1)	Correct cube roots for negative inputs