Traditional Abacus and Bead Arithmetic/Roots/Cube root

Theory

Let $x$ be the number of which we want to obtain the cube root $y={\sqrt[{3}]{x}}$ ; Let's consider its decimal expansion, for example: $x=456.7890123$. Let's separate its digits into groups of three around the decimal point in the following way

$$x=456.7890123=456.789\cdot 012\cdot 300\cdot 000$$

or, in other words, let's define the sequence of integers $\alpha _{i}$

$$\alpha _{1}=456,\alpha _{2}=789,\alpha _{3}=012,\alpha _{4}=300,\alpha _{5}=000\cdots$$

and let's build the sequence $x_{i}$ recursively from $x_{0}=0$

$$x_{i}=1000x_{i-1}+\alpha _{i}$$

and let $y_{i}$ be the integer part of the cube root of $x_{i}$

$$y_{i}=\lfloor {\sqrt[{3}]{x}}_{i}\rfloor$$

i.e.  $y_{i}$ is the largest integer whose cube is not greater than $x_{i}$. Finally, let us call remainders to the differences

$$r_{i}=x_{i}-y_{i}^{3}\geq 0$$

For our example we have:

$i$	$\alpha _{i}$	$x_{i}$	$y_{i}$	$r_{i}$
0		0	0	0
1	456	456	7	113
2	789	456789	77	256
3	012	456789012	770	256012
4	300	456789012300	7701	78119199
5	000	456789012300000	77014	6949021256
⋯

Let's see that, by construction, $x_{i}$ grows as $10^{3i}$ (three more digits in each step), in fact the sequence $x_{i}10^{3-3i}$, i.e. 0, 400, 456, 456.789, 456.789012, etc. tends to $x$ ($x_{i}10^{3-3i}\to x$). By comparison, $y_{i}$ , as the integer part of the cube root of $x_{i}$, grows only as $10^{i}$ (one digit more each step). As $y_{i}$ is the largest integer whose square is not greater than $x_{i}$ we have $r_{i}=xi-y_{i}^{3}\geq 0$ as above but

$$x_{i}-(y_{i}+1)^{3}=x_{i}-y_{i}^{3}-3y_{i}^{2}-3y_{i}-1<0$$

by definition of $y_{i}$, or

$$r_{i}=x_{i}-y_{i}^{3}<3y_{i}^{2}+3y_{i}+1$$

multiplying by $10^{3-3i}$

$$x_{i}\cdot 10^{3-3i}-(y_{i}\cdot 10^{1-i})^{3}<(3y_{i}^{2}+3yi+1)\cdot 10^{3-3i}$$

but as  $y_{i}$ grows only as $10^{i}$, the second term tends to zero as $10^{-i}$.  

$$x_{i}\cdot 10^{3-3i}-(y_{i}\cdot 10^{1-i})^{3}\to 0$$

and $x_{i}\cdot 10^{3-3i}\to x$ so that we have

$$y_{i}\cdot 10^{1-i}\to y={\sqrt[{3}]{x}}$$

For other numbers, the above factors are: $10^{3k-3i}$ and $10^{k-i}$, where $k$ is the number of three-digit groups to the left of the decimal point, negative if it is followed by 000 groups (ex. $k=0$ for $x=0.00456$, $k=-2$ for $x=0.000000456$, etc.).

This is the basis for traditional manual cube root methods.

Procedure

We start with $i=0,x_{0}=0,y_{0}=0,r_{0}=0$.

First digit

$b$	$b^{3}$
1	1
2	8
3	27
4	64
5	125
6	216
7	343
8	512
9	729

For $i=1,x_{1}=\alpha _{1}$. It is trivial to find $y_{1}$ such that its square does not exceed $x_{1}$ through the use of the following table of cubes that can be easily retained in memory. In the case of the example it is $y_{1}=7$

Rest of digits

For $i>1$, we have $x_{i}=1000x_{i-1}+\alpha _{i}$ as defined above and we try to build $y_{i}$ in the way:

$$y_{i}=10y_{i-1}+b$$

where $b$ is a one-digit integer ranging from 0 to 9. To obtain $b$ we have to choose the greatest digit from 0 to 9 such that:

$$x_{i}-y_{i}^{3}=x_{i}-(10y_{i-1}+b)^{3}\geq 0$$

or

$$x_{i}-(a+b)^{3}\geq 0$$

if we write $a=10y_{i-1}$. Expanding the binomial we have

$$x_{i}-a^{3}-3a^{2}b-3ab^{2}-b^{3}=1000x_{i-1}+\alpha _{i}-(10y_{i-1})^{3}-3a^{2}b-3ab^{2}-b^{3}\geq 0$$

or

$$1000r_{i-1}+\alpha _{i}\geq 3a^{2}b+3ab^{2}+b^{3}=(3a^{2}+3ab+b^{2})b$$

The left side of the above expression may be seen as simply the previous remainder with the next three-digits group appended to it. If we evaluate the term on the right for each value of $b$ and compare with the left term we have:

$a=10y_{i-1}=70$

$b$	$(3a^{2}+3ab+b^{2})b$	$1000r_{i-1}+\alpha _{i}$
0	0	≤ 113789
1	14911	≤ 113789
2	30248	≤ 113789
3	46017	≤ 113789
4	62224	≤ 113789
5	78875	≤ 113789
6	95976	≤ 113789
7	113533	≤ 113789	⬅
8	131552	> 113789
9	150039	> 113789

and it is clear that the next figure of our root is a 7 but, how can we proceed in the general case without having to systematically explore every possibility ($b=0,1,\cdots ,9$)?

Here Knott distinguishes two different approaches:

• Preparing the divisor
• Preparing the dividend

Preparing the divisor

This correspond with the above expression

$$1000r_{i-1}+\alpha _{i}\geq 3a^{2}b+3ab^{2}+b^{3}=(3a^{2}+3ab+b^{2})b$$

And it is the strategy usually used with paper and pencil and can also be implemented, of course, on the abacus. In the above expression, if we see the left part as dividend and the parentese as divisor, $b$ is the first digit of the division

$$b=(1000r_{i-1}+\alpha _{i})/(3a^{2}+3ab+b^{2})$$

but since we don't know b yet, we approximate it using only the main part of the divisor

$$b=(1000r_{i-1}+\alpha _{i})/(3a^{2})=(1000r_{i-1}+\alpha _{i})/(300y_{i-1}^{2})$$

This gives us a guess as to what the value of $b$ might be, but we need:

1. Verify that the value thus obtained is correct, or, where appropriate, correct it up or down as needed.
2. Obtain the next remainder to prepare the calculation of the next digit of the root.

You can see an example in Tone nikki blog, see also Modern approaches below.

Preparing the dividend

Starting again with

$$1000r_{i-1}+\alpha _{i}\geq (3a2+3ab+b2)b=3a\left(a+b+{\frac {b}{3a^{2}}}\right)b$$

we prepare the dividend by dividing $1000r_{i-1}+\alpha _{i}$ (the next three-digits group appended to the previous remainder) by $3a$

$$(1000r_{i-1}+\alpha _{i})/3a\geq \left(a+b+{\frac {b}{3a^{2}}}\right)b$$

As usual, we don't know $b$ and we can't evaluate the parentheses on the right, but we can get a clue about $b$ by approximating the parentheses by its main part $a$ and use it as a trial divisor.

$$\left(a+b+{\frac {b}{3a^{2}}}\right)\approx a$$

so that

$$b\approx (1000r_{i-1}+\alpha _{i})/3a^{2}$$

After that, we need again:

1. Verify that the value thus obtained is correct, or, where appropriate, correct it up or down as needed.
2. Obtain the next remainder to prepare for the obtention of the next digit of the root by evaluating $1000r_{i-1}+\alpha _{i}-3a^{2}b-3ab^{2}-b^{3}$.

Please note that:

• Divisor 3 is involved in the prepared dividend and this leads to non-finite decimal fractions.
• The division by $a$ not only worsens the above, but also makes the prepared dividend specific to the current step, since the value of $a$ evolves with the calculation of the different figures of the result.

This did not occur in the calculation of square roots and, as a consequence, the process of obtaining cube roots is much more complicated and requires a complex cycle of preparation-restoration of the dividend that, following Knott, can be represented by the following scheme:

• a) Divide by $a$.
• b) Divide by 3.
• c) Obtain $b$ as the first digit of the division of the above by $a$.
• d) Subtract $b(a+b)$ (Equivalent to subtracting $3a^{2}b$ and $3ab^{2}$ in $1000r_{i-1}+\alpha _{i}-3a^{2}b-3ab^{2}-b^{3}$).
• e) Multiply by 3.
• f) Multiply by $a$.
• g) Subtract $b^{3}$.

In our example ($x=456.7890123$), using traditional division and traditional division arrangement (like Knott does), working the two first digits:

Cube root of 456.7890123

Abacus	Comment
ABCDEFG
456789	Enter number aligning first group with B
-343	-7^3=343
113789	First remainder
7113789	Enter 7 in A as first root digit and append second group
7113789	a) Divide B-F by 7
7162554	b) Divide B-F by 3
7541835	c) Divide B by A (one digit)
7751835	d) Subtract 7*7=49 from CD
77 2835	e) Multiply CDEF by 3. Add 3✕283 to CDEFG
77  854	f) Multiply CDEF by 7. Add 7✕85 to CDEFG
77  599
-343	g) Subtract 7^3=343 to CDEFG
77  256	New remainder
...	Root obtained so far: 7.7

Notes:

• ^1 a) It is unnecessary to extend the division by 7 beyond the current three-digit group. The 4 in G is a division remainder meaning 4/7.
• ^2 b) The same can be said of division by 3. It is carried out up to column F and the remainder (1) is temporarily added to column G. The value (5) in said column is a strange hybrid meaning 1/3 and 4/7 . It does not matter, it will be reabsorbed in steps e) and f).

Modern approaches

Members of the Soroban & Abacus Group modified the technique described by Knott to adapt it to modern soroban use. The result is allegedly faster at the expense of being less compact and requiring an abacus with more rods to store intermediate data. The simplicity of having the result directly substituting the radicand is also lost.

You can also find a compilation of modern methods for both square and cube roots in Tone Nikki (とね日記) by a Japanese blogger (Author's name does not appear to be available).

Examples of cube roots

The following examples are all worked using traditional division and traditional division arrangement. Components of the dividend preparation-restoration cycle are labelled with a), b), etc as detailed above.

Cube root of 157464

Abacus	Comment
ABCDEFG	Cube root of 157464
157464	Enter number aligning first group with B
-125	Subtract 5^3=125 from BCD
32464	First remainder: 32
5 32464	Enter 5 in A as first root digit and append second group
5 32464	a) Divide C-F by 5 (G will be the division remainder)
5 64924	b) Divide C-F by 3
5216404	c) Divide B by 5
5416404	d) Subtract 4x4=16 from CD
54  404	e) Multiply 40x3 in EFG (adding to remainder in G)
54  124	f) Multiply 12x5 in EFG
54   64	g) Subtract 4^3=64 from FG
54	Remainder 0; Done! Root is 54

Clearly, if the remainder is zero and there are no more (not null) groups to add, the number is a perfect cube and we are done. Root is 54.

Cube root of 830584

Abacus	Comment
ABCDEFG	Cube root of 830584
830584	Enter number aligning first group with B
-729	Subtract 9^3=729 from BCD
101584	101: first remainder
9101584	Enter 9 in A as first root digit and append second group
9101584	a) Divide C-F by 9 (G will be the division remainder)
9112871	b) Divide C-F by 3
9376232	c) Divide B by 9 (A)
9416232	d) Subtract 4x4=16 from CD
94  232	e) Multiply 23x3 in EFG (adding to remainder in G)
94   71	f) Multiply 07x9 in EFG
94   64	g) Subtract 4^3= 64 from FG
94	Remainder 0; Done! Root is 94

Root is 94.

Cube root of 666

Abacus	Comment
ABCDEFG	Cube root of 666
666	Enter 666 in BCD
+	(Unit rod)
-512	Subtract 8^3=512 from BCD
154	First remainder
8154	Enter 8 in A as the first root digit
8154000	Append 000 as new group
8154000	a) Divide B-F by 8 (A)
8192500	b) Divide B-F by 3
8641662	c) Divide B by 8 (A)
8781662	d) Subtract BxB=49 from CD
8732662	e) Multiply C-F by 3 in C-G
87 9800	f) Multiply C-F by 8 (A) in C-G
87 7840	g) Subtract B^3=343 from EFG
87 7497	Root so far 8.7, Remainder 7.497

Now we continue using Abbreviated operations. We need to divide the remainder (7497) by three times the square of the current root ($3\cdot 87^{2}=22707$)

Abacus	Comment
ABCDEFGHIJKLM
87 7497
87 7497------	Squaring 87
+49	7^2
+112	2*7*8
+64	8^2
87 7497  7569	multiplying by 3 (adding double)
+14
+10
+12
+18
87 7497 22707	dividing 7497/22707, two digits
...
8733	Root 8.733 (Compare to: ${\sqrt[{3}]{666}}=8.7328917$)

Cube root of 237176659 (three digits)

Abacus	Comment
ABCDEFGHIJ	Cube root of 237176659
237176659	Enter number aligning first group with B
-216	Subtract 6^3=216 from BCD
21176659	21: first remainder
21176659	Enter 6 in A as first root digit and append second group
6 21176659	a) Divide B-F by 6 (A)
6 35292659	b) Divide B-F by 3
6117633659	c) Divide B by 6 (A)
6157633659	d) Subtract BxB=1 from CD
6156633659	e) Multiply C-F by 3 in C-G
6116992659	f) Multiply C-F by 8 (A) in C-G
6110196659	g) Subtract B^3=343 from EFG
6110195659	Root so far 61, Remainder 10195
----------
6110195659	Append third group
6110195659	a) Divide C-H by 61 (AB)
6116714158	b) Divide C-H 3
6155713678	c) Divide C by 61 (AB)
6190813678	d) Subtract CxC=81 from EF
619   3678	e) Multiply D-H by 3 in D-I
619   1158	f) Multiply D-H by 61 (AB) in D-J
619    729	g) Subtract C^3=729 from HIJ
619    000	Done, no remainder!
----------	Root is 619

Cube root of $48^{3}-1=110591$ to eight digits

The first triplet 110 is between 64 and 125, so that the cube root of 110 591 is between 40 and 50. First root digit is 4

First digit:

Cube root of 110591: First digit

Abacus	Comment
ABCDEFG	Cube root of 110591
110591	Enter number aligning first group with B
-64	Subtract 6^3=216 from BCD
46591	46: first remainder
46591	Enter 4 in A as first root digit and append second group
4 46591	OK 1st digit!

Second digit:

Cube root of 110591: Second digit

Abacus	Comment
ABCDEFG
4 46591	a) Divide B-F by 4 (A)
4116473	b) Divide B-F by 3
4388234	c) Divide B by 4 (A)
4868234	d) Subtract BxB=64 from CD
48 4234	e) Multiply C-F by 3 in C-G
48 1273	f) Multiply C-F by 4 (A) in C-G
48  511	g) Cannot subtract 8^3=512 from EFG! Going back (See note at the end)
48  511	-f) Divide C-F by 4 (A)
48 1273	-e) Divide C-F by 3
48 4234	-d) Add 8x8=64 in CD
4868234	-c) Revise down B
-1
+4
47T8234	d) Subtract BxB=49 from CD (T=10)
4759234	e) Multiply C-F by 3 in C-G
4717773	f) Multiply C-F by 4 (A) in C-G
47 7111	g) Subtract B^3=343 from EFG
47 6768	OK 2nd digit! Remainder 6768

Third digit:

Cube root of 110591: Third digit

Abacus	Comment
ABCDEFGHIJ
47 6768000	Append 000 to previous remainder
47 6768000	a) Divide C-H by 47 (AB)
4714400000	b) Divide C-H 3
4748000000	c) Divide C by 47 (AB)
4795700000	d) Subtract C^2=81 from EF
4794890000	e) Multiply D-H by 3 in D-I
4792298300	f) Multiply D-H by 47 (AB) in D-J
479 689490	g) Subtract C^3=729 from HIJ
479 688761	OK 3rd digit! Remainder 688761

Fourth digit:

Cube root of 110591: Fourth digit

Abacus	Comment
ABCDEFGHIJKLM
479 688761000	Append 000 to previous remainder
479 688761000	a) Divide D-J by 479
4791437914194	b) Divide D-J by 3
4794793046394	c) Divide D by 479 1d
4799482046394	d) Subtract 9^2=81 from GH
4799473946394	e) Multiply E-J by 3 in E-K
4799142184194	f) Multiply E-J by 479 in E-M
4799 68106330	g) Subtract -D^3=729 from KLM
4799 68105601	Ok 4th digit! Remainder 68105601

Now we finish the calculation using abbreviated operations. We need to divide the remainder (68105601) by three times the square of the current root (4799). The first four digits of the result are appended after the ones already obtained; for instance:

Cube root of 110591: Continue using abbreviated operations

Abacus	Comment
ABCDEFGHIJKLM
4799 68105601	Divide E-M by 4799
479914191623	Divide E-M by 4799
47992957204	Divide E-M by 3
47999857	Compare this to ${\sqrt[{3}]{110591}}=47.9998553236\cdots$

As we can see, we have obtained a result with 7 correct figures.

Note: We found above that with root 48 we could not subtract $8^{3}=512$, or we had a negative remainder (-1). This might seem unfortunate since it forced us to undo part of the work and correct the new root figure downwards, but in practice what we find is a fortunate result: the small remainder (-1) tells us that 48 was a excellent approximation (by excess) to the root, opening a new way to solve the problem. In fact, what we have is:

$$110591=48^{3}-1=48^{3}\left(1-{\frac {1}{48^{3}}}\right)$$

or

$${\sqrt[{3}]{110591}}=48{\sqrt[{3}]{1-{\frac {1}{48^{3}}}}}$$

where we can use

$${\sqrt[{3}]{1-a}}\approx 1-{\frac {a}{3}}$$

so that

$${\sqrt[{3}]{110591}}=48{\sqrt[{3}]{1-{\frac {1}{48^{3}}}}}\approx 48\left(1-{\frac {1}{3\cdot 48^{3}}}\right)=48-{\frac {1}{3\cdot 48^{2}}}=47.999855324$$

compare to ${\sqrt[{3}]{110591}}=47.999855323638$. We could have thus achieved great precision with little effort!

From elementary arithmetic to numerical analysis

The abacus is currently studied as a traditional art or as a means to develop numerical and cognitive skills in general, it is not expected that in the computer age it will be used as a calculator to solve real world problems. But if that were the case and you had to solve a large number of cube roots (something unusual) you might want to move from traditional methods or basic arithmetic to modern numerical analysis methods and try the Newton-Raphson method. You can find an adaptation of this method to the abacus in jccAbacus...

Appendix: Cubes of two digits numbers

Cubes of two-digits numbers

	+1	+2	+3	+4	+5	+6	+7	+8	+9
10	1331	1728	2197	2744	3375	4096	4913	5832	6859
20	9261	10648	12167	13824	15625	17576	19683	21952	24389
30	29791	32768	35937	39304	42875	46656	50653	54872	59319
40	68921	74088	79507	85184	91125	97336	103823	110592	117649
50	132651	140608	148877	157464	166375	175616	185193	195112	205379
60	226981	238328	250047	262144	274625	287496	300763	314432	328509
70	357911	373248	389017	405224	421875	438976	456533	474552	493039
80	531441	551368	571787	592704	614125	636056	658503	681472	704969
90	753571	778688	804357	830584	857375	884736	912673	941192	970299

This can help you practice two-digit cube roots.

Example: ${\sqrt[{3}]{250047}}=60+3=63$