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Arithmetic Algebra: Cube Roots

Main Arithmetic Algebra Page

Page Contents

  1. Rationale
  2. Application to algebra
  3. Application to Arithmetic
    1. An Example which works as expected.
    2. An Example Which Needs Fine Tuning
    3. A Longer Example Which Requires Fine Tuning

Rationale

Noting that:cubeRoots1.gif[2.1]

We try to understand how to extract the cube root:

We seek: cubeRoots2.gif [2.2]
We know the answers in the following and it is the method which is important. Also, this gives a vocabulary when considering applications in algebra and arithmetic.

a+b Answer
cubeRoots3.gif Cube
a3 We choose a as the cube root a
3a2b 3a2b+3ab2+b3 Remainder
To find b, we can use (3a2+3ab+b2)b, to divide into the remainder to find b.
Or assuming that 3a2b is the largest term and the rest is small in comparison, then we can divide by 3a2, to find b.
 a+b
(3a2+3ab+b2)b Remainder after using (a+b) as the root so far:
(3a2+3ab+b2)b
0

Observations:
  1. First we seek a cube root, a. 
  2. Having chosen it, we subtract a3 from the cube
  3. With the remainder, we need to find a number, b, such that 3a2b+3ab2+b3≤remainder
  4. We can estimate b by dividing the remainder by 3a2.
  5. This division gives us b+b/a+b3/3a2, if b is small in respect to a, then we get a fair estimate of b, otherwise, not. 
  6. In arithmetic, where a and b are numbers between 0 and 10, we actually have b+b/(10a)+b3/300a2, which makes it more likely that the b estimated by dividing by 3a2 is close to the real b. 

Application to algebra


Answer: cubeRoots4.gif Answer so far
1+x Find a
a=1		 1
1 Subtract a3 from the cube
3·(x/3) x Remainder after subtracting a3
Find b, so 3b≤-x
b=x/3		 1+x/3
x+x2/3+x3/27 3a2b+3ab2+b3
a=1, b=x/3
=x+x2/3+x3/27
(3+2x+x2/3)(-x2/9) -x2/3-x3/27 Remainder.
New a=1+x/3
a2=1+2x/3+x2/9
3a2=3+2x+x2/3
b=-x2/27		 1+x/3-x2/9
(3+2x+x2/3)·(-x2/9)+
(1+x/3)·(x4/81)+
x6/729		 Subtract:
3a2b+3ab2+b3=
(3+2x+x2/3)·(-x2/9)+
(1+x/3)·(x4/81)+
x6/729
3·5x3/81... (-x3/27)+2x3/9....
=5x3/27...		 Remainder.
New a=(1+x/3-x2/9)
...

Below is the expansion of (1+x)1/3, to term 10:
cubeRoots5.gif [2.3]

Application to Arithmetic

In applying the method to arithmetic, we note that instead of our remainder being 3a2b+3ab2+b3, it is:

300a2b+30ab2+b3

Where a and b are numbers between 0 and 10.

So to estimate b, we divide the remainder:
300a2
Using this approximation there are sometimes problems with the second figure of the cube, and fine tuning is required, see the second example. This occurs when the first figure is low, 1 for example, and the next one is high, 9, for example.

If this estimate ( 300a2) does not give a number between 0 and 10, we can fine tune it to:
300b+30b2+b3 ≤ remainder
Divide throughout by 300:
b+b2/10+b3/300 ≤ remainder/300
This is the exact expression, but the arithmetic is very much easier after the division.

Also, because a number between 0 and 10, when cubed, is less than 1000, we divide the cube into groups of 3.

An Example Which Works As Expected.

We seek the cube root of 362467097
This example, has been carefully engineered to work as expected. That is, the second figure of the cube is small compared with the first.

Answer: 713 Answer so far
362 467 097 We diviide the number in groups of three. 7
343 We find the nearest cube less than 362, which is 73=343. Subtract this from the cube.
14 700·b 19 467 Remainder.
a=7, 300a2=14 700
Estimate b as 19467/14700, say 1		 71
14 911 Calculate 300a2b+30ab2+b3
300a2b=14700
30ab2=210
b3=1
1512300·b 4 556 097 Subtract 300a2b+30ab2+b3, and bring down the rest of the number
a=71,
300a2=300.712=1512300
Estimate b as: 4556097/1512300
Estimate is 3.
 713
4556097 Compute 300a2b+30ab2+b3=
1512300+
19170+
27
The remainder is zero, so we have our cube root.

An Example Which Needs Fine Tuning

Sometimes, it isn't easy to find the second figure. The problem and solution are explained below in the example.
We seek the cube root of 6859.

Answer 1 9 Answer so far
6 859
1  a=1, we subtract a3 1
5 859 b is estimated as:
5859/300
Giving b as approximately 19. (b must be between 0 and 10).
We need to be more precise to find b.
300b+30b2+b3≤5859
Divide throughout by 300:
b+b2/10+b3/300≤19.53
Try b=9, approximately:
9+81/10+729/300≏19
As this is very close, we will work this out precisely:
9+8.1+2.43=19.53
As this is exactly the figure we require, we have finished.

19

A Longer Example Which Requires Fine Tuning

We seek the cube root of 5 079 577 959
Answer 1719 Answer so far
5 079 577 959
1 The nearest cube less than 5 is 1 1
4 079 300b≤4079, from this, b is approximately 13, but it must be between 0 and 10.
So we have no idea what b might be!
We need to fine tune:
300b+30b2+b3≤4079
Divide throughout by 300:
b+b2/10+b3/300≤13.593..
Try b=9: (working approximately)
9+8+2=19
This is too big, so try:
b=8:
8+6+1=15
Also too big.
Try b=7
7+5+1=13
Working precisely now:
7+49/10+342/300=12.006...
So b=7 is  correct.
 17
3913 300·7+30·72+73=3913
We subtract this number:
166 577 a=17,
300a2=86700
86700b≤166577
b≅1,

171
87211 Take b=1, so
300·172+30·17+1=
86700+510+1=
87211
Subtract this from the remainder so far.
7 9366 959 Bring down the next group of 3.
a=171
300·1712=8772300
8772300b≅79366959
79366959 b≏9
300a2b=78950700
30ab2=415530
b3=729
Subtract the total		 1719
0 As this leaves zero, we are done.

It seems that when there is a problem with this method, it is with the second figure, when it isn't small in comparison to the first one. That is, b isn't small compared with a.

By being more precise, and working approximately, it is much easier to find the value for b. After this, the normal rule of 300a2b as an approximation for b seems to work fine, because the new a's are always large compared with the new b's and the longer the computation, the more this is true, so b becomes smaller and smaller in relation to a.










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