▪ Shane Baggs method
▪ Steve Treadwell's modified method

ABACUS: MYSTERY OF THE BEAD
The Bead Unbaffled - An Abacus Manual


Abacus Arithmetic

by Welton J. Crook

Cube Root

The extraction of the cube root of a number on the abacus is more complicated but not more tedious than the extraction of the square root. However, it can be accomplished and the method is of interest, perhaps, from this point of view. As far as practicability is concerned, the use of logarithmic tables is much to be preferred.

The general procedure for extracting cube root is as follows:

  1. Place the "cube" number on the right side of the abacus and point it off, starting at the decimal point, into groups each containing three numerals.

  2. Place 1, on column one, on the left hand side of the abacus. This is the "root" number.

  3. Place 1 on column six (from the left) and call it the "square" number. We now have three sets of numbers, from left to right, the "root" number, the "square" number and the "cube" number.

  4. Subtract the 1 of the "square" number from the first or left hand group of the "cube" number.

  5. Add 1 to the "root" number and add the sum to the "square" number.

  6. Add an additional 2 to the "root" number and add the sum to the "square" number.

  7. Subtract the sum of the "square" number from the residue of the "cube" number and repeat the process given in *(f) and (g) until the number in the first group of the "cube" number is less than the "square" number. *A group member points out a typo. The instructions should read "repeat the process given in (e) (f) and (g)"  (see corrected examples).

  8. Now attach the second group (three numerals) of the cube number, to the residue of the first group. Add 1 to the "root" number and add the sum of the "root" number to the square number. Then add 0 to the "root" number and add 11. **Add the sum (total root number) to the "square" number but on the second column to the right, from the left hand figure of the "square" number. For example: If the total "root" number is 61 and the square number is 12, the new square number will be 1261.

  9. Subtract the new "square" number from the "cube" number and proceed as before until the "cube" number is exhausted.

  10. If, after attaching a new group of the "cube" number, the "square" number is still too large, attach an additional group but instead of adding 0 to the "root" number, followed by the addition of 11, in this case add 00 followed by the addition of 101. ***The sum is then added to the "square" number on the fourth column to the right from its left hand figure.

  11. When the "cube" number has been exhausted, add 2 to the final "root" number and divide by 3. The quotient is the cube root sought.

** *** Those most familiar with the technique feel that some procedures, as written by Crook, could confuse students and very well lead to an incorrect answer. With this in mind several group members have recommended changes to clarify sections h and j. (Please see changes.)

If the number, whose cube root is sought, consists of five or more numerals, an abacus with more than thirteen columns will be required. For instance, when extracting the cube root from a number containing five numerals, the columns occupied on the abacus will be:

By root number    3
By square number    4
By cube number    5
__
Total    12

With a 13-column abacus, there will not be enough vacant columns between the various sets of numbers. This would lead to confusion.

Example I

  3____
 \/9261 = 21
  1. Place 9261 on the right side of the abacus. This is the "cube" number. Place 1 on column one on the left hand side of the frame. This is the "root" number. Place 1 on a column near the center of the abacus, say, on column six. This is the "square" number.

  2. Point off the "cube" number, starting at the decimal, into groups containing three numerals. In the case of 9261, the first or left hand group will consist of the number 9. The second group will consist of the numbers 261.

  3. Subtract the "root" number, 1, (from first group of cube number) leaving 8.

  4. Add 1 to the "root" number (1 + 1 = 2) and add this to the 1 of the "square" number (1 + 2 = 3). Change one on column six from the left to 3. The "root" number is now 2. The "square" number is now 3 and the "cube" number is 8 (on column four from the right).

  5. Add 2 to the "root" number, 2 + 2 = 4. Add the sum of the root number, 4, to the "square" number, 3 + 4 = 7. This makes 7 on column six from the left. Subtract the sum of the "square" number, namely 7, from the "cube" number, 8 - 7 = 1, leaving 1 on column four from the right. The "square" number is now larger than the residue from the first group of the "cube" number.

  6. Attach the second group of the "cube" number to the residue of the first group. The "cube" number is now 1261. Add 1 to the "root" number, 4 + 1 = 5. Add the sum of the "root" number, namely 5, to the "square" number, which is 7.   5 + 7 = 12.

  7. Add 0 to the "root" number, making 50 and add 11, making 61. This is the new "root" number. Attach the numerals 6 and 1 to the "square" number to form 1261. The situation on the abacus is shown in Figure 34, in which, numbering the columns from the left, the "root" number, 61, occupies columns one and two; the "square" number, 1261, occupies columns five, six, seven and eight; the "cube" number, 1261, occupies columns ten, eleven, twelve and thirteen.   
     

    Fig. 34. Example I in cube root. Condition on the abacus after operation (g) when extracting the cube root of 9261.

  8. Subtract the "square" number from the "cube" number.   1261 - 1261 = 0. There remains on the abacus only the "root" number, namely 61. Add 2 to the "root" number and divide by 3.   61 + 2 = 63.   63 ÷ 3 = 21.   21 is the cube root sought.