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

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2. Find Cube root of 216 using Babylonian method, with Initial guess x0 = 5

Solution:
Babylonian method for finding cube roots by hand
`:. root (3)(216)=?`

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

`x_0=5`


`1^(st)` iteration :

`x_1=(2*5+216/5)/3`

`=(10+8.64)/3`

`=(18.64)/3`

`=6.2133`


`2^(nd)` iteration :

`x_2=(2*6.2133+216/6.2133)/3`

`=(12.4267+5.5951)/3`

`=(18.0217)/3`

`=6.0072`


`3^(rd)` iteration :

`x_3=(2*6.0072+216/6.0072)/3`

`=(12.0145+5.9855)/3`

`=(18)/3`

`=6`


`4^(th)` iteration :

`x_4=(2*6+216/6)/3`

`=(12+6)/3`

`=(18)/3`

`=6`

`:. root (3)(216)=6`

Babylonian Method Table

Iteration	`x`	`216/x`	Average
1	5	8.64	6.2133
2	6.2133	5.5951	6.0072
3	6.0072	5.9855	6
4	6	6	6

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3. Find Cube root of 2 using Babylonian method, with Initial guess x0 = 1

Solution:
Babylonian method for finding cube roots by hand
`:. root (3)(2)=?`

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

`x_0=1`


`1^(st)` iteration :

`x_1=(2*1+2/1)/3`

`=(2+2)/3`

`=(4)/3`

`=1.3333`


`2^(nd)` iteration :

`x_2=(2*1.3333+2/1.3333)/3`

`=(2.6667+1.125)/3`

`=(3.7917)/3`

`=1.2639`


`3^(rd)` iteration :

`x_3=(2*1.2639+2/1.2639)/3`

`=(2.5278+1.252)/3`

`=(3.7798)/3`

`=1.2599`


`4^(th)` iteration :

`x_4=(2*1.2599+2/1.2599)/3`

`=(2.5199+1.2599)/3`

`=(3.7798)/3`

`=1.2599`

`:. root (3)(2)=1.2599`

Babylonian Method Table

Iteration	`x`	`2/x`	Average
1	1	2	1.3333
2	1.3333	1.125	1.2639
3	1.2639	1.252	1.2599
4	1.2599	1.2599	1.2599

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4. Find Cube root of 5 using Babylonian method, with Initial guess x0 = 2

Solution:
Babylonian method for finding cube roots by hand
`:. root (3)(5)=?`

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

`x_0=2`


`1^(st)` iteration :

`x_1=(2*2+5/2)/3`

`=(4+1.25)/3`

`=(5.25)/3`

`=1.75`


`2^(nd)` iteration :

`x_2=(2*1.75+5/1.75)/3`

`=(3.5+1.6327)/3`

`=(5.1327)/3`

`=1.7109`


`3^(rd)` iteration :

`x_3=(2*1.7109+5/1.7109)/3`

`=(3.4218+1.7082)/3`

`=(5.1299)/3`

`=1.71`


`4^(th)` iteration :

`x_4=(2*1.71+5/1.71)/3`

`=(3.42+1.71)/3`

`=(5.1299)/3`

`=1.71`

`:. root (3)(5)=1.71`

Babylonian Method Table

Iteration	`x`	`5/x`	Average
1	2	1.25	1.75
2	1.75	1.6327	1.7109
3	1.7109	1.7082	1.71
4	1.71	1.71	1.71

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