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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 will 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)

Beneath is the implementation of above thought.

## 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)

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