# Python Program to find the cube of each list element

• Difficulty Level : Basic
• Last Updated : 17 Feb, 2021

Given a list, the task is to write a python program to cube all the list elements.

Input: [1, 2, 3, 4]

Output: [1, 8, 27, 64]

Explanation: Cubing all the list elements

Input: [2, 4, 6]

Output: [8, 64, 216]

Method 1 : Using loop

This is the brute force way. In this, we just multiply the same element two times to itself.

Example:

## Python3

 `# Initializing list``l ``=` `[``1``, ``2``, ``3``, ``4``]`` ` `# Cube List using loop``res ``=` `[]``for` `i ``in` `l:``    ``res.append(i``*``i``*``i)`` ` `# printing result``print``(res)`

Output:

[1, 8, 27, 64]

Method 2 : Using pow() function

This is also the brute force way. In this, we use in-built pow() function

Example:

## Python3

 `# Initializing list``l ``=` `[``1``, ``2``, ``3``, ``4``]`` ` `# Cube List using loop``res ``=` `[]``for` `i ``in` `l:``    ``res.append(``pow``(i, ``3``))`` ` `# printing result``print``(res)`

Output:

[1, 8, 27, 64]

Method 3 : Using list comprehension

This task can also be performed using list comprehension. This is similar to above function. Just the difference is that its compact and one liner.

Example:

## Python3

 `# Initializing list``l ``=` `[``1``, ``2``, ``3``, ``4``]`` ` `# Cube List using list comprehension``res ``=` `[``pow``(i, ``3``) ``for` `i ``in` `l]`` ` `# printing result``print``(res)`

Output:

[1, 8, 27, 64]

Method 4: Using lambda

This can also be achieved using lambda function

Example:

## Python3

 `# Initializing list``l ``=` `[``1``, ``2``, ``3``, ``4``]`` ` `res ``=` `list``(``map``(``lambda` `x: x ``*``*` `3``, l))``print``(res)`

Output:

[1, 8, 27, 64]

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