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# Python Program for Discover cubic root of a quantity

Given a quantity n, discover the dice root of n.
Examples:

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

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

We are able to use binary search. First we outline error e. Allow us to say 0.0000001 in our case. The principle steps of our algorithm for calculating the cubic root of a quantity n are:

1. Initialize begin = 0 and finish = n
2. Calculate mid = (begin + finish)/2
3. Test if absolutely the worth of (n – mid*mid*mid)
4. If (mid*mid*mid)>n then set finish=mid
5. If (mid*mid*mid)

Under is the implementation of above concept.

## Python3

 ` `  `def` `diff(n, mid) :` `    ``if` `(n > (mid ``*` `mid ``*` `mid)) :` `        ``return` `(n ``-` `(mid ``*` `mid ``*` `mid))` `    ``else` `:` `        ``return` `((mid ``*` `mid ``*` `mid) ``-` `n)` `         `  `def` `cubicRoot(n) :` `     `  `    ` `    ` `    ``begin ``=` `0` `    ``finish ``=` `n` `     `  `    ` `    ``e ``=` `0.0000001` `    ``whereas` `(``True``) :` `         `  `        ``mid ``=` `(begin ``+` `finish) ``/` `2` `        ``error ``=` `diff(n, mid)` ` `  `        ` `        ` `        ` `        ``if` `(error <``=` `e) :` `            ``return` `mid` `             `  `        ` `        ` `        ``if` `((mid ``*` `mid ``*` `mid) > n) :` `            ``finish ``=` `mid` `             `  `        ` `        ` `        ``else` `:` `            ``begin ``=` `mid` `             `  `n ``=` `3` `print``(``"Cubic root of"``, n, ``"is"``, ` `      ``spherical``(cubicRoot(n),``6``))`

Output:

`Cubic root of three.000000 is 1.442250`

Time Complexity: O(logn)

Auxiliary Area: O(1)