Ken Ward's Mathematics Pages

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:[2.1]

We try to understand how to extract the cube root:

We seek: [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
		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:			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:
[2.3]
As with the square root, the expansion of the cube root gives us a pre-Binomial way of expanding expressions.

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 divide 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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Faster Arithmetic - by Ken Ward