SOME methods of finding the cube roots of arithmetical quantities shall be given, but since the operations in most cases are attended with much labour, it will be better, when required, to find the cube roots and all other higher roots by means of logarithms.

Sometimes, however, it is requisite to obtain an approximate root: thus if the cube root of 612 were required, we

know the cube of 8 to be 512, and of 9 to be 729; assume therefore the cube root of 612 to be 8.5, and 8.53=614·125, which is very near, and the root may therefore be assumed to be 8.47 in most cases a little reflection will enable any one to approximate sufficiently near the root for small quantities.

TO FIND THE CUBE ROOT.

Since the cube of any number will not be made up of more than 3 digits, divide the number into periods of three, beginning at units, and find the greatest cube number in the first period on the left hand, and set its root in the quotient for the first figure of the required root.

Subtract the cube from the period above it, and bring down the next period to the remainder for a dividend.

Divide the dividend by 300 times the square of the figure in the root, and the quotient figure will be the second figure in the root:

Subtract the cube of the two figures in the root from the two first periods on the left hand, and to the remainder bring down the next period for a new dividend.

Divide this dividend by 300 times the square of the two figures, and the quotient figure is the third figure in the

root:

Subtract the cube of the three figures in the root from the three left hand periods; then proceed as before till all the periods are brought down.

N. B. Should there be a remainder after all the figures of the proposed number are brought down, periods of 3 ciphers each may be annexed, and the root continued in decimals.

divisor 1702 × 300 = 8670000) 60940243 ( 7

Here we first divide the number into periods, by placing the dot over the last figure, and then over every third figure

beginning from it. We take the cube number next less than the first period 80; this is 64, and we set its cube root, 4, as the first figure in the root; we then subtract the 64, and bring down the next period 677. We now triple the first figure of the root, and set it at some distance to the left of the remainder; the result is 12 (there is 123 in the sum, but the 3 will be accounted for by and by); then we multiply this triple by the first figure of the root, and place the product 48 between the 12 and the remainder, adding two ciphers to it.

We now divide the remainder by this 4800, and get the second figure in the root 3, which we also set after the 12, making 123: now we multiply this 123 by 3, the second figure in the root, and set the product 369 under 4800, add them up, and multiply the sum 5169 by the second figure in the root, subtracting the product 15507 from the remainder. The remainder is 1170, to which we bring down the next period 568, and have now to form the two quantities to the left of it. The first is obtained by tripling the last figure 3 of 123 which gives 129 (the final 2 in 1292 will be accounted for when the next figure in the root is found); and the other quantity 5547 is found by adding 9, the square of the second

figure in the root, to the two preceding middle lines

369

5169

We have now only to add two ciphers, and repeat the whole process described in this paragraph.

To find the Cube Roots of the same Numbers by Logarithms.

And 2.2322323= |1707, which is at once as near as by all the tedious operation; and the same method is applicable to all higher roots or powers, and since logarithms are now so commonly used, there is little judgment in seeking after laborious methods, when there exists such a quick and

LOGARITHMS are such indices of any number, that the said number being raised successively to those powers will express any and all numbers in succession, and this number is termed the base of the system.

Any number might be used, but the number 10 was adopted by Briggs, and is practically the most useful. Thus if 2 were the base, we should have

20 = 1, 2 = 2, 2 = 4, 23 = 8, 24 = 16, &c.

=

and there might be found some indices between 2 and 3, such that 2 raised to those powers might represent 5, 6, or 7, &c. Again, 100: = 1, 101 = 10, 102 = 100, 103 = 1000, 104= 10000; and then again there can be found such indices of 10