# 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 = 10Output:33.1%Input:x = 50Output: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; ` `} ` |

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## 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.. ` |

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## 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 ` |

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## 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.. ` |

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**Output:**

percentage increase in the volume of the cube is 33.1%

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