Cubes and Cube Roots

Calculating the cube root of integer perfect cubes

For perfect cubes, it is possible to calculate the root by hand.

For n3, where n is an integer, the method of extracting the root is as follows.

  A B Root Comments
a     n0=n3 Place n3under Root, call this n0
b 1   n1 = n0 - 1 Place 1 under A and subtract from Root, new Root is now n1
c   1   Place 1 under Root_Number_B
d1 2 3   Add 1 to A, add to last B value
e1 4 7 n2 = n1 - 7 Add 2 to A, add to last B value, now subtract new B from Root, new root is now n2
d1 5 12   Add 1 to A, add to last B value
e1 7 19 n3 = n2 - 19 Add 2 to A, add to last B value, now subtract new B from Root, new root is now n3

Repeat the process d and e until root n = 0. The value of the Root_Number e when n = 0 is then added to 2 and the sum is divided by 3 which should give the cube root of n^3.

A real example follows.

4x4x4 = 64, so the cube root of 64 is 4.

A B Root
    64
1   63
  1  
2 3  
4 7 56
5 12  
7 19 37
8 27  
10 37 0

The value of A when Root = 0 is 10. Adding 2 to 10 and dividing by 3 = 12/3 = 4

Series

The numbers which are subtracted from the Root form a series

Sn = 1 + 7 + 19 + 37 + .....

If one looks at the sum, the number of terms relates to the cube root of the sum itself.

The expression for n3 sum can be in the form

Where

(1st term) 1 = 1
(2nd term) 7 = 1 + 6
(3rd term) 19 = 1 + 6 + 12
(4th term) 37 = 1 + 6 + 12 + 18
.
.

By inspection, we see that each successive term contains 1 + a number of terms values which are multiples of 6. Since the first term is 1, the last term is the 6.(n-1).


(nth term) = 1 + 6 + 12 + 18 + 24 + + 6.(n-1)

Inspecting the terms, we find that

The term 6.(n-1) will appear n-(n-1) times or just once.

Thus the sum becomes

The mathematical summations indicated by the sigma signs are standard identities [Tuckey & Armistead, p.211]. So, from the generalised equation of the summation for

Sn = 1 + 7 + 19 + 37 + ...... + un

and identified a

we have found it to be equivalent to the cube of n.

That is,

Sn = n + 6[ (n-1) + 2(n-2) + 3(n-3) + ... + (n-1)(n-[n-1]) ] = n3

Reference


Index

© Dylan Sung 2005

This page was created on Thursday 2nd June 2005