All numbers expressed by 1, 2, or 3 digits have for cube roots numbers expressed by 1 digit. Next we have the following table:

Thus all numbers expressed by 4, 5, or 6 digits have for cube roots numbers expressed by 2 digits; further, the first digit of the root in such case is determined by the first one (to the left), the first two, or the first three digits of the number, according as it is expressed by 4, 5, or 6 digits. Thus the cube roots of the numbers. 2744, 39304, 357911, will in every case be numbers expressible by two digits. The tens digits will be determined by the 2, the 39, the 357 of the numbers to be 1, 3, 7 respectively.

To find the cube root of a number we shall see how the cube of a number is formed. The identity of the following two ways of multiplying 14 by 14, and the product by 14, will at once be seen:

We see, then, that the cube of such a number as 14that is, a number regarded as being made up of two parts, here 10 and 4—is the cube of one part increased by three times the product of the square of that part and the other part, and three times the product of the first part and the square of the second, and the cube of the second part.

We wish now to recover from 2744 its cube root. Plainly, the tens digit of the root is 1-that is, the first part of the root is ten; take from 2744 the cube of 10: 2744|10 1000

1744

The remainder is made up of three parts:

(1) The product of 3 times the square of 10, and the other part of the root.

(2) The product of 3 times 10, and the square of the other part of the root.

(3) The cube of the other part of the root.

Then, if we divide 1744 by 3 times the square of 10, we shall have a clue to the other part of the root. Dividing, we may take 5 as the other part:

Taking away 3 X 10' X 5, we have as remainder 244, which should be made up of parts (2) and (3), mentioned above. We find, however, that it is not large enough. We have taken a second part too large, and therefore take a smaller part, say 4:

=

Here the remainder is 544, which 3 X 10 X 4' + 4', and we conclude that the root is 14.

We see also that the 1744 = 4 (3 × 62 + 4×3× 10+4'). The work might then be shown thus:

The further development of the method will follow lines similar to those followed in square root. We shall take space only to indicate how, in the case of finding a cube root consisting of several figures, a certain saving of work may be secured:

When we reach the point where we wish to determine the third figure, we have to find three times the square of 93-that is,

3 (90+2 × 90 × 3+3')

Now, as indicated above, 810 is 3 x 90 x 3, 9 is 3*, 25119 is 3X 90+3 x 90 x 3+3', so that, if to the sum of 810, 9, 25119 we add 9, which is the square of

3, we shall have found three times the square of 93. If to the resulting number we affix two zeros, we shall have three hundred times the square of 93.

This artifice may be employed when at each successive stage we need three hundred times the square of the part already found.

We shall conclude this chapter with the remark that the fourth root of a number is to be found by extracting the square root of its square root, and the sixth root of a number by extracting the square root of its cube root.

THE END.

(15)