# 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:
* Verify that the value thus obtained is correct, or, where appropriate, correct it up or down as needed.
* 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:
* Verify that the value thus obtained is correct, or, where appropriate, correct it up or down as needed.
* 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: ${\sqrt[{3}]{250047}}=60+3=63$

## References

* ↑ Knott, Cargill G. (1886), "The Abacus, in its Historic and Scientific Aspects", Transactions of the Asiatic Society of Japan, 14: 18–73
* 1 2 Tone? (2017). "Square root and Cube root using Abacus". とね日記. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)
* ↑ 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)
* ↑ Cabrera, Jesús (2021). "Newton's method for abacus; square, cubic and fifth roots". jccAbacus. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)