1. Find Cube root of 4096 using Babylonian method, with Initial guess x0 = 15

Solution:
Babylonian method for finding cube roots by hand
`:. root (3)(4096)=?`

`x_(i+1)=(2*x_i+s/(x_i)^2)/3`

`x_0=15`


`1^(st)` iteration :

`x_1=(2*15+4096/15)/3`

`=(30+18.2044)/3`

`=(48.2044)/3`

`=16.0681`


`2^(nd)` iteration :

`x_2=(2*16.0681+4096/16.0681)/3`

`=(32.1363+15.8646)/3`

`=(48.0009)/3`

`=16.0003`


`3^(rd)` iteration :

`x_3=(2*16.0003+4096/16.0003)/3`

`=(32.0006+15.9994)/3`

`=(48)/3`

`=16`


`4^(th)` iteration :

`x_4=(2*16+4096/16)/3`

`=(32+16)/3`

`=(48)/3`

`=16`

`:. root (3)(4096)=16`

Babylonian Method Table

Iteration	`x`	`4096/x`	Average
1	15	18.2044	16.0681
2	16.0681	15.8646	16.0003
3	16.0003	15.9994	16
4	16	16	16