# Program to evaluate the expression (√X+1)^6 + (√X-1)^6

Given a number . The task is to find the value of the below expression for the given value of . Examples:

Input: X = √2
Output: 198
Explanation: = 198

Input: X = 3
Output: 4160

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to use Binomial expression. We can take these two terms as 2 binomial expressions. By expanding these terms we can find the desired sum. Below is the expansion of the terms. Now put X= in EQ(1) Below is the implementation of above approach:

## C++

 // CPP program to evaluate the given expression  #include  using namespace std;     // Function to find the sum  float calculateSum(float n)  {      int a = int(n);         return 2 * (pow(n, 6) + 15 * pow(n, 4)               + 15 * pow(n, 2) + 1);  }     // Driver Code  int main()  {      float n = 1.4142;         cout << ceil(calculateSum(n)) << endl;         return 0;  }

## Java

 // Java program to evaluate the given expression  import java.util.*;     class gfg  {  // Function to find the sum  public static double calculateSum(double n)  {      return 2 * (Math.pow(n, 6) + 15 * Math.pow(n, 4)               + 15 * Math.pow(n, 2) + 1);  }     // Driver Code  public static void main(String[] args)  {      double n = 1.4142;      System.out.println((int)Math.ceil(calculateSum(n)));  }  }  //This code is contributed by mits

## Python3

 # Python3 program to evaluate   # the given expression     import math     #Function to find the sum  def calculateSum(n):             a = int(n)             return (2 * (pow(n, 6) + 15 * pow(n, 4)               + 15 * pow(n, 2) + 1))         #Driver Code  if __name__=='__main__':      n = 1.4142     print(math.ceil(calculateSum(n)))     # this code is contributed by   # Shashank_Sharma

## C#

 // C# program to evaluate the given expression  using System;  class gfg  {  // Function to find the sum  public static double calculateSum(double n)  {      return 2 * (Math.Pow(n, 6) + 15 * Math.Pow(n, 4)               + 15 * Math.Pow(n, 2) + 1);  }     // Driver Code  public static int Main()  {      double n = 1.4142;      Console.WriteLine(Math.Ceiling(calculateSum(n)));      return 0;  }  }  //This code is contributed by Soumik

## PHP

 

Output:

198


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