Percentage increase in volume of the cube if a side of cube is increased by a given percentage

Given here is a cube, whose one side is increased by a given percentage. The task is to find percentage increase in the volume of the cube.
Examples:

Input: x = 10
Output: 33.1%

Input: x = 50
Output: 237.5%

Approach

• In a cube, all sides are equal, so, 
  length = breadth = height
• let side of the cube = a
• given percentage increase = x%
• so, volume before increase = a^3
• after increase, new side = a + ax/100
• so, new volume = (a + ax/100)^3 = a^3 + (ax/100)^3 + 3a^3x/100 + 3a^3x^2/10000
• increase in volume = new volume – old volume = (a^3 + (ax/100)^3 + 3a^3x/100 + 3a^3x^2/10000) – a^3 = (ax/100)^3 + 3a^3x/100 + 3a^3x^2/10000
• so, percentage increase in volume = (((ax/100)^3 + 3a^3x/100 + 3a^3x^2/10000)/a^3) * 100 = ((x/100)^3 + 3x/100 + 3x^2/10000) * 100 = x^3/10000 + 3x + 3x^2/100

Below is the implementation of the above approach: 

C++

// C++ program to find percentage increase
// in the volume of the cube
// if a side of cube is increased
// by a given percentage

#include <bits/stdc++.h>
using namespace std;

void newvol(double x)
{
    cout << "percentage increase "
         << "in the volume of the cube is "
         << pow(x, 3) / 10000 + 3 * x
                + (3 * pow(x, 2)) / 100
         << "%" << endl;
}

// Driver code
int main()
{
    double x = 10;
    newvol(x);
    return 0;
}

Java

// Java program to find percentage increase
// in the volume of the cube
// if a side of cube is increased
// by a given percentage
import java.io.*;

class GFG
{

static void newvol(double x)
{
    System.out.print( "percentage increase "
    +"in the volume of the cube is "
        + (Math.pow(x, 3) / 10000 + 3 * x
                + (3 * Math.pow(x, 2)) / 100) );
                System.out.print("%");
}

// Driver code
public static void main (String[] args)
{
    double x = 10;
    newvol(x);
}
}

// This code is contributed by anuj_67..

Python3

# Python program to find percentage increase
# in the volume of the cube
# if a side of cube is increased
# by a given percentage

def newvol(x):

    print("percentage increase"
            "in the volume of the cube is ",
            ((x**(3)) / 10000 + 3 * x
                + (3 * (x**(2))) / 100),"%");

x = 10;
newvol(x);

# This code is contributed by PrinciRaj1992

C#

// C# program to find percentage increase
// in the volume of the cube
// if a side of cube is increased
// by a given percentage
using System;

class GFG
{

static void newvol(double x)
{
    Console.Write( "percentage increase "
    +"in the volume of the cube is "
        + (Math.Pow(x, 3) / 10000 + 3 * x
                + (3 * Math.Pow(x, 2)) / 100) );
                Console.Write("%");
}

// Driver code
public static void Main ()
{
    double x = 10;
    newvol(x);
}
}

// This code is contributed by anuj_67..

Javascript

<script>
// javascript program to find percentage increase
// in the volume of the cube
// if a side of cube is increased
// by a given percentage
function newvol( x)
{
    document.write("percentage increase "
         + "in the volume of the cube is "
         + (Math.pow(x, 3) / 10000 + 3 * x
                + (3 * Math.pow(x, 2)) / 100)
         + "%" );
}

// Driver code
    let x = 10;
    newvol(x);

// This code is contributed by gauravrajput1
</script>

Output:

percentage increase in the volume of the cube is 33.1%

Time Complexity: O(1) because pow function will take constant time

Auxiliary Space: O(1), since no extra space has been taken.