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
a^{3}
We choose a as the cube root
a
3a^{2}b
3a^{2}b+3ab^{2}+b^{3}
Remainder
To find b, we can use (3a^{2}+3ab+b^{2})b,
to divide into the remainder to find b.
Or assuming that 3a^{2}b is the largest term and
the rest is small in comparison, then we can divide by 3a^{2},
to find b.
a+b
(3a^{2}+3ab+b^{2})b
Remainder after using (a+b)
as the root so far:
(3a^{2}+3ab+b^{2})b
0
Observations:
First we seek a cube root, a.
Having chosen it, we subtract a^{3}
from the cube
With the remainder, we need to find a number, b, such that
3a^{2}b+3ab^{2}+b^{3}≤remainder
We can estimate b by dividing the remainder by 3a^{2}.
This division gives us b+b/a+b^{3}/3a^{2},
if b is small in respect to a, then we get a fair estimate of b,
otherwise, not.
In arithmetic, where a and b are numbers between 0 and 10,
we actually have b+b/(10a)+b^{3}/300a^{2},
which makes it more likely that the b estimated by dividing by 3a^{2}
is close to the real b.
Application
to algebra
Answer:
Answer so far
1+x
Find a
a=1
1
1
Subtract a^{3} from the cube
3·(x/3)
x
Remainder after subtracting a^{3}
Find b, so 3b≤-x
b=x/3
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 3a^{2}b+3ab^{2}+b^{3},
it is:
300a^{2}b+30ab^{2}+b^{3}
Where a and b are numbers between 0 and 10.
So to estimate b, we divide the remainder:
300a^{2}
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 (
300a^{2}) does not give a number between 0 and 10, we can fine tune it to:
300b+30b^{2}+b^{3} ≤ remainder
Divide throughout by 300:
b+b^{2}/10+b^{3}/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 7^{3}=343.
Subtract this from the cube.
14 700·b
19 467
Remainder.
a=7, 300a^{2}=14 700
Estimate b as 19467/14700, say 1
Subtract 300a^{2}b+30ab^{2}+b^{3},
and bring down the rest of the number
a=71,
300a^{2}=300.71^{2}=1512300
Estimate b as: 4556097/1512300
Estimate is 3.
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 a^{3}
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+30b^{2}+b^{3}≤5859
Divide throughout by 300:
b+b^{2}/10+b^{3}/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+30b^{2}+b^{3}≤4079 Divide throughout by 300: b+b^{2}/10+b^{3}/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+30·7^{2}+7^{3}=3913 We subtract this number:
166 577
a=17, 300a^{2}=86700 86700b≤166577 b≅1,
171
87211
Take b=1, so 300·17^{2}+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=171 300·171^{2}=8772300 8772300b≅79366959
79366959
b≏9 300a^{2}b=78950700 30ab^{2}=415530 b^{3}=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 300a^{2}b
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.
Ken's book is packed with examples and explanations that enable you to discover more than 150 techniques to speed up your arithmetic and increase your understanding of numbers. Paperback and Kindle: