II. We obtain the second term of the 1st column by adding the first term to itself; the result being multiplied

by this first term, and added to column, gives its second term. term to the second term of the third term.

the first term of the 2d Again, adding this first 1st column, we get its

III. We seek how many times the second term of the 2d column is contained in the first dividend; or, simply how many times it is contained in its first part, 24000, which gives 5 for the second part of the root.

IV. Finally, we add this 5 to the last term of the 1st column, whose result, multiplied by 5, and added to the last term of the 2d column, gives its third term; which, multiplied by 5, gives 27125=24000+3000+125.

This work can be written in a more condensed form, as follows, where the ciphers upon the right have been omitted.

Case I.

From the preceding operation, we may draw the following rule for extracting the cube root of a whole number.

RULE.

I. Since the cube of any number cannot have more than three times as many places of figures as the number, we must separate the number into periods of three figures each, counting from the unit's place towards the left. When the number of figures is not divisible by 3, the left-hand period will contain less than three figures.

II. Seek the greatest cube of the first, or left-hand period; place its root at the right of the number, after the manner of a quotient in division; also place it to the left of the number for the first term of a column marked 1st COL. Then multiply it into itself, and place the product for the first term of a column marked 2d COL. Again, multiply this last result by the same figure, and subtract the product from the first period, and to the remainder annex the next period, and it will give the FIRST DIVIDEND. This same figure must be added to the first term of the 1st column; the sum will be its second term, which must be multiplied by the same figure, and the product added to the first term of the 2d column ; this sum will be its second term, which we shall name the

FIRST TRIAL DIVISOR.

The same figure of the root must be added to the second term of the 1st column, to form its third term.

III. See how many times the trial divisor, with two

ciphers annexed, is contained in the dividend; the quotient figure will be the second figure of the root, which must be placed at the right of the first figure; also annex it to the third term of the 1st column, and multiply the result by this second figure, and add the product, after advancing it two places to the right, to the last term of the 2d column. Again, multiply this last result by this second figure of the root, and subtract the product from the dividend, and to the remainder annex the next period for a NEw dividend.

Proceed with this second figure of the root precisely as was done with the first figure, and so continue, until all the periods have been brought down.

NOTE. This rule may be readily deduced from the rule for extracting the cube root of a polynomial, as given in my ALGEBRA.

The greatest cube of the first period, 387, is 343, whose root is 7, which we place to the right of the number for the first figure of the root sought. We also

place it for the first term of the first column, which, multiplied into itself, gives 7x7=49, for the first term of the 2d column, which, in turn, multiplied by 7, gives 49×7=343, which, subtracted from the first period, 387, leaves the remainder 44, to which, annexing the next period, 420, we get 44420 for the first dividend.

Again, adding 7 to the first term, 7, of the 1st column, we get 7+7=14, for the second term of the 1st column, which, multiplied by 7, gives 14x7=98; this, added to the first term of the second column, gives 147 for the second term of the 2d column, or the first trial divisor.

Again, adding 7 to the second term of the 1st column, we get 14+7=21, for the third term of the first column.

The trial divisor, with two ciphers annexed, becomes 14700, which is contained 3 times in the first dividend, 44420. Since the trial divisor is less than the true divisor, it will sometimes give too large a quotient figure; such is the case in this present example, where 2 is the second figure of the root.

This second figure, 2, of the root, annexed to the third term of the 1st column, gives 212; which, multiplied by 2, gives 424, which, being advanced two places to the right, must be added to 147, the last term of the 2d column. The sum 15124 will form the third term of the 2d column, which, multiplied by 2, gives 15124 x2=30248, which, subtracted from the first dividend, leaves 14172 for the remainder, to which, annexing the next period, 489, we get 14172489 for the second dividend.

Again, to the last term, 212, of the 1st column, adding 2, we get 214 for the next term; which, multiplied by

2, gives 428, which, added to 15124, gives 15552 for the second trial divisor. Again, adding 2 to 214, we get 216 for the fifth term of the 1st column.

The second trial divisor, with two ciphers annexed, becomes 1555200, which is contained 9 times in the second dividend, 14172489; therefore, 9 is the third figure of the root, which, annexed to 216, gives 2169 for the last term of the first column, which, multiplied by 9, gives 19521, which, advanced two places to the right, and then added to 15552, gives 1574721; this, multiplied by 9, gives 14172489, which, subtracted from the second dividend, leaves no remainder.