We next calculate the third or last figure of the root.
As the first figure of the second period of the cube is so large, it will be unsafe to disregard it. Call the first period, therefore, 596·9; all other figures may be neglected.
| 596·9 | mill. | ||
| (2) | 8³ = | 512 | ” |
| 84·9 | ” | ||
| (3) deduct 84 x 8 x 4 x 3 = (roughly) | 80·6 | ” | |
| (5) divide by 84² x 3 = (88 x 80 x 3)[378] = 2·1 | 4·3 | ||
| 2 | |||
| Quotient—2, which is the third or last figure of the root. | |||
[Note.—I have not encumbered the above figures with the ciphers which should accompany them, as, to the expert calculator, this will be needless.]
The root, therefore, is 842.
It is stated in the text that my pupils could extract the cube roots of numbers ranging as high as 2,000,000,000. In the ordinary mode this number would be divided, as above, into four periods; but my pupils treated the 2,000 as one period, the approximate root of which is of course 12, the cube of 12 being 1,728.