# Find cubic root of a number

• Difficulty Level : Medium
• Last Updated : 01 Sep, 2022

Given a number n, find the cube root of n.
Examples:

```Input:  n = 3
Output: Cubic Root is 1.442250

Input: n = 8
Output: Cubic Root is 2.000000```

Recommended Practice

We can use binary search. First we define error e. Let us say 0.0000001 in our case. The main steps of our algorithm for calculating the cubic root of a number n are:

1. Initialize start = 0 and end = n
2. Calculate mid = (start + end)/2
3. Check if the absolute value of (n – mid*mid*mid) < e. If this condition holds true then mid is our answer so return mid.

4. If (mid*mid*mid)>n then set end=mid
5. If (mid*mid*mid)<n set start=mid.

Below is the implementation of above idea.

## C++

 `// C++ program to find cubic root of a number``// using Binary Search``#include ``using` `namespace` `std;` `// Returns the absolute value of n-mid*mid*mid``double` `diff(``double` `n,``double` `mid)``{``    ``if` `(n > (mid*mid*mid))``        ``return` `(n-(mid*mid*mid));``    ``else``        ``return` `((mid*mid*mid) - n);``}` `// Returns cube root of a no n``double` `cubicRoot(``double` `n)``{``    ``// Set start and end for binary search``    ``double` `start = 0, end = n;` `    ``// Set precision``    ``double` `e = 0.0000001;` `    ``while` `(``true``)``    ``{``        ``double` `mid = (start + end)/2;``        ``double` `error = diff(n, mid);` `        ``// If error is less than e then mid is``        ``// our answer so return mid``        ``if` `(error <= e)``            ``return` `mid;` `        ``// If mid*mid*mid is greater than n set``        ``// end = mid``        ``if` `((mid*mid*mid) > n)``            ``end = mid;` `        ``// If mid*mid*mid is less than n set``        ``// start = mid``        ``else``            ``start = mid;``    ``}``}` `// Driver code``int` `main()``{``    ``double` `n = 3;``    ``printf``(``"Cubic root of %lf is %lf\n"``,``           ``n, cubicRoot(n));``    ``return` `0;``}`

## Java

 `// Java program to find cubic root of a number``// using Binary Search``import` `java.io.*;` `class` `GFG``{``    ``// Returns the absolute value of n-mid*mid*mid``    ``static` `double` `diff(``double` `n,``double` `mid)``    ``{``        ``if` `(n > (mid*mid*mid))``            ``return` `(n-(mid*mid*mid));``        ``else``            ``return` `((mid*mid*mid) - n);``    ``}``    ` `    ``// Returns cube root of a no n``    ``static` `double` `cubicRoot(``double` `n)``    ``{``        ``// Set start and end for binary search``        ``double` `start = ``0``, end = n;`` ` `        ``// Set precision``        ``double` `e = ``0.0000001``;`` ` `        ``while` `(``true``)``        ``{``            ``double` `mid = (start + end)/``2``;``            ``double` `error = diff(n, mid);`` ` `            ``// If error is less than e then mid is``            ``// our answer so return mid``            ``if` `(error <= e)``                ``return` `mid;`` ` `            ``// If mid*mid*mid is greater than n set``            ``// end = mid``            ``if` `((mid*mid*mid) > n)``                ``end = mid;`` ` `            ``// If mid*mid*mid is less than n set``            ``// start = mid``            ``else``                ``start = mid;``        ``}``    ``}``    ` `    ``// Driver program to test above function``    ``public` `static` `void` `main (String[] args)``    ``{``        ``double` `n = ``3``;``        ``System.out.println(``"Cube root of "``+n+``" is "``+cubicRoot(n));``    ``}``}` `// This code is contributed by Pramod Kumar`

## Python3

 `# Python 3 program to find cubic root``# of a number using Binary Search` `# Returns the absolute value of``# n-mid*mid*mid``def` `diff(n, mid) :``    ``if` `(n > (mid ``*` `mid ``*` `mid)) :``        ``return` `(n ``-` `(mid ``*` `mid ``*` `mid))``    ``else` `:``        ``return` `((mid ``*` `mid ``*` `mid) ``-` `n)``        ` `# Returns cube root of a no n``def` `cubicRoot(n) :``    ` `    ``# Set start and end for binary``    ``# search``    ``start ``=` `0``    ``end ``=` `n``    ` `    ``# Set precision``    ``e ``=` `0.0000001``    ``while` `(``True``) :``        ` `        ``mid ``=` `(start ``+` `end) ``/` `2``        ``error ``=` `diff(n, mid)` `        ``# If error is less than e``        ``# then mid is our answer``        ``# so return mid``        ``if` `(error <``=` `e) :``            ``return` `mid``            ` `        ``# If mid*mid*mid is greater``        ``# than n set end = mid``        ``if` `((mid ``*` `mid ``*` `mid) > n) :``            ``end ``=` `mid``            ` `        ``# If mid*mid*mid is less``        ``# than n set start = mid``        ``else` `:``            ``start ``=` `mid``            ` `# Driver code``n ``=` `3``print``(``"Cubic root of"``, n, ``"is"``,``      ``round``(cubicRoot(n),``6``))`  `# This code is contributed by Nikita Tiwari.`

## C#

 `// C# program to find cubic root``// of a number using Binary Search``using` `System;` `class` `GFG {``    ` `    ``// Returns the absolute value``    ``// of n - mid * mid * mid``    ``static` `double` `diff(``double` `n, ``double` `mid)``    ``{``        ``if` `(n > (mid * mid * mid))``            ``return` `(n-(mid * mid * mid));``        ``else``            ``return` `((mid * mid * mid) - n);``    ``}``    ` `    ``// Returns cube root of a no. n``    ``static` `double` `cubicRoot(``double` `n)``    ``{``        ` `        ``// Set start and end for``        ``// binary search``        ``double` `start = 0, end = n;` `        ``// Set precision``        ``double` `e = 0.0000001;` `        ``while` `(``true``)``        ``{``            ``double` `mid = (start + end) / 2;``            ``double` `error = diff(n, mid);` `            ``// If error is less than e then``            ``// mid is our answer so return mid``            ``if` `(error <= e)``                ``return` `mid;` `            ``// If mid * mid * mid is greater``            ``// than n set end = mid``            ``if` `((mid * mid * mid) > n)``                ``end = mid;` `            ``// If mid*mid*mid is less than``            ``// n set start = mid``            ``else``                ``start = mid;``        ``}``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main ()``    ``{``        ``double` `n = 3;``        ``Console.Write(``"Cube root of "``+ n``                       ``+ ``" is "``+cubicRoot(n));``    ``}``}` `// This code is contributed by nitin mittal.`

## PHP

 ` (``\$mid` `* ``\$mid` `* ``\$mid``))``        ``return` `(``\$n` `- (``\$mid` `*``                ``\$mid` `* ``\$mid``));``    ``else``        ``return` `((``\$mid` `* ``\$mid` `*``                 ``\$mid``) - ``\$n``);``}` `// Returns cube root of a no n``function` `cubicRoot(``\$n``)``{``    ` `    ``// Set start and end``    ``// for binary search``    ``\$start` `= 0;``    ``\$end` `= ``\$n``;` `    ``// Set precision``    ``\$e` `= 0.0000001;` `    ``while` `(true)``    ``{``        ``\$mid` `= ((``\$start` `+ ``\$end``)/2);``        ``\$error` `= diff(``\$n``, ``\$mid``);` `        ``// If error is less``        ``// than e then mid is``        ``// our answer so return mid``        ``if` `(``\$error` `<= ``\$e``)``            ``return` `\$mid``;` `        ``// If mid*mid*mid is``        ``// greater than n set``        ``// end = mid``        ``if` `((``\$mid` `* ``\$mid` `* ``\$mid``) > ``\$n``)``            ``\$end` `= ``\$mid``;` `        ``// If mid*mid*mid is``        ``// less than n set``        ``// start = mid``        ``else``            ``\$start` `= ``\$mid``;``    ``}``}` `    ``// Driver Code``    ``\$n` `= 3;``    ``echo``(``"Cubic root of \$n is "``);``    ``echo``(cubicRoot(``\$n``));` `// This code is contributed by nitin mittal.``?>`

## Javascript

 ``

Output:

`Cubic root of 3.000000 is 1.442250`

Time Complexity: O(logn)

Auxiliary Space: O(1)
This article is contributed by Madhur Modi .If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

My Personal Notes arrow_drop_up