# Traditional Abacus and Bead Arithmetic/Roots/Cube root

## Theory

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

${\displaystyle x=456.7890123=456.789\cdot 012\cdot 300\cdot 000}$

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

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

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

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

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

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

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

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

For our example we have:

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

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

by definition of ${\displaystyle y_{i}}$, or

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

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

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

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

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

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

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

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

This is the basis for traditional manual cube root methods.

## Procedure

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

### First digit

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

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

### Rest of digits

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

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

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

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

or

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

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

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

or

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

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 ${\displaystyle b}$ and compare with the left term we have:

${\displaystyle a=10y_{i-1}=70}$
${\displaystyle b}$ ${\displaystyle (3a^{2}+3ab+b^{2})b}$ ${\displaystyle 1000r_{i-1}+\alpha _{i}}$
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 (${\displaystyle b=0,1,\cdots ,9}$)?

Here Knott[1] distinguishes two different approaches:

• Preparing the divisor
• Preparing the dividend

#### Preparing the divisor

This correspond with the above expression

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

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, ${\displaystyle b}$ is the first digit of the division

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

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

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

This gives us a guess as to what the value of ${\displaystyle b}$ 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[2], see also Modern approaches below.

#### Preparing the dividend

Starting again with

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

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

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

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

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

so that

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

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 ${\displaystyle 1000r_{i-1}+\alpha _{i}-3a^{2}b-3ab^{2}-b^{3}}$.

• Divisor 3 is involved in the prepared dividend and this leads to non-finite decimal fractions.
• The division by ${\displaystyle a}$ not only worsens the above, but also makes the prepared dividend specific to the current step, since the value of ${\displaystyle a}$ 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 ${\displaystyle a}$.
• b) Divide by 3.
• c) Obtain ${\displaystyle b}$ as the first digit of the division of the above by ${\displaystyle a}$.
• d) Subtract ${\displaystyle b(a+b)}$ (Equivalent to subtracting ${\displaystyle 3a^{2}b}$ and ${\displaystyle 3ab^{2}}$ in ${\displaystyle 1000r_{i-1}+\alpha _{i}-3a^{2}b-3ab^{2}-b^{3}}$).
• e) Multiply by 3.
• f) Multiply by ${\displaystyle a}$.
• g) Subtract ${\displaystyle b^{3}}$.

In our example (${\displaystyle x=456.7890123}$), 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 71
7162554 b) Divide B-F by 32
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[3]. 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)[2] 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 (${\displaystyle 3\cdot 87^{2}=22707}$)

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: ${\displaystyle {\sqrt[{3}]{666}}=8.7328917}$)

### 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 ${\displaystyle 48^{3}-1=110591}$ 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 ${\displaystyle {\sqrt[{3}]{110591}}=47.9998553236\cdots }$

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 ${\displaystyle 8^{3}=512}$, 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:

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

or

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

where we can use

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

so that

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

compare to ${\displaystyle {\sqrt[{3}]{110591}}=47.999855323638}$. 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[4]...

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

## References

1. Knott, Cargill G. (1886), "The Abacus, in its Historic and Scientific Aspects", Transactions of the Asiatic Society of Japan, 14: 18–73
2. a b Tone? (2017). "Square root and Cube root using Abacus". とね日記. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)
3. Baggs, Shane; Heffelfinger, Totton (2011). "Cube Roots". 算盤 Abacus: Mystery of the Bead. Archived from the original on August 1, 2021. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)
4. Cabrera, Jesús (2021). "Newton's method for abacus; square, cubic and fifth roots". jccAbacus. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)

Next Page: Abbreviated operations | Previous Page: Square root