Calculating the cube root of integer perfect cubes
For perfect cubes, it is possible to calculate the root by hand.
For n^{3}, where n is an integer, the method of extracting the root is as follows.
A B Root Comments a n_{0}=n^{3} Place n^{3}under Root, call this n_{0} b 1 n_{1} = n_{0} - 1 Place 1 under A and subtract from Root, new Root is now n_{1} c 1 Place 1 under Root_Number_B d1 2 3 Add 1 to A, add to last B value e1 4 7 n_{2} = n_{1} - 7 Add 2 to A, add to last B value, now subtract new B from Root, new root is now n_{2} d1 5 12 Add 1 to A, add to last B value e1 7 19 n_{3} = n_{2} - 19 Add 2 to A, add to last B value, now subtract new B from Root, new root is now n_{3} 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
S_{n} = 1 + 7 + 19 + 37 + .....
If one looks at the sum, the number of terms relates to the cube root of the sum itself.
1^{3} term 1 = 1
2^{3} term 8 = 1 + 7
3^{3} term 27 = 1 + 7 + 19
4^{3} term 64 = 1 + 7 + 19 + 37
.
.
n^{3} term = 1 + 7 + 19 + 37 + …… + (u_{n})The expression for n^{3} 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
- 1 occurs n times
- 6 = 6.1 occurs n-1 times
- 12 = 6.2 occurs n-2 times
- 18 = 6.3 occurs n-3 times
- 24 = 6.4 occurs n-4 times
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
S_{n} = 1 + 7 + 19 + 37 + ...... + u_{n}
and identified a
we have found it to be equivalent to the cube of n.
That is,
S_{n} = n + 6[ (n-1) + 2(n-2) + 3(n-3) + ... + (n-1)(n-[n-1]) ] = n^{3}
Reference
© Dylan Sung 2005
This page was created on Thursday 2nd June 2005