part of the root becomes gradually larger, and the unknown part gradually smaller; that, every time a new root-figure is obtained, the given number is diminished by the cube of the part of the root then known; that the resulting remainder consists principally of the product of 3 times the square of the known by the unknown part of the root; and that, consequently, the division of the remainder by 3 times the square of the known part of the root must give the rest of the root, or a number not much larger-according as the remainder does or does not. contain (a) 3 times the product of the known part of the root by the square of the quotient, and (b) the cube of the quotient, in addition to 3 times the product of the square of the known part by the quotient.

In the contracted process, we merely dispense with unnecessary ciphers, and (instead of writing the remainders in full) bring down the figures according as they are required—that is, in "periods" of three each: no use being made of the last three figures until the tens' figure of the root has been determined; no use being made of the last six figures until the hundreds' figure of the root has been determined; and so on, as we proceed to the left. (See § 226.) When, therefore, we place à dot over every third figure, beginning with the units' figure, we are able at once to see what particular period is to be brought down at each stage of the work: we are also able to tell, beforehand, how many places the root will occupya place for every dot.

238. To extract the Cube Root of a (whole) number: Divide the figures into periods of three each, by placing a dot over every third figure, beginning with the units' figure; and should either one figure or two remain on the extreme left, regard this portion of the number as a period also. Find the highest figure whose cube is contained in the most left-hand period, and set it down as the most left-hand figure of the required root. Subtract the cube of this figure from the first period, and to the remainder annex the second period: the number thus formed will be the first "dividend.” Convert this dividend into a "trial"-dividend, by conceiving the last two figures cut off; and, for "trial-divisor," take 3 times the square of the root-figure. Divide the trial-dividend by the trial-divisor, and set down the resulting quotient-figure, upon trial, as the second figure of the

root. Then, to 3 times the first root-figure annex the quotient-figure, and multiply the resulting number by the quotient-figure; to the product add the trial-divisor-removed two places to the left; and multiply the sum by the quotient-figure. If the number so obtained be contained in the dividend, retain the quotient-figure as the second figure of the root; if not, try a lower quotient-figure. Having found, and written as the second rootfigure, the highest figure which, employed in the manner just described, gives a result not greater than the dividend, subtract the result from the dividend, and to the remainder annex the third period: the number so formed will be the second dividend. Convert this new dividend into a new trial-dividend, by conceiving the last two figures cut off; and, for a new trial-divisor, take 3 times. the square of the part of the root now known. Treat this pair of numbers-the new trial-dividend and trial-divisor-in the same manner as the preceding pair; and continue the process so long as any period remains to be brought down. The result so obtained will be the cube root of the given number-when the last subtraction leaves no remainder; and will, when there is a remainder, be the cube root of the difference between the given number and this remainder.

Thus, in the case of the last example (contracted process), we begin by dividing the figures of the given number into five periods the two figures (15) on the extreme left being regarded as a period. For the first root-figure, we set down 2, whose cube (8) is the largest contained in 15. Subtracting 23, or 8, from 15, and bringing down the next period (783), we have, as the first "dividend," 7783. Conceiving the last two figures (83) of this dividend cut off, we obtain the first “trial”dividend, 77. For "trial-divisor," we treble the square of the first root-figure: 22 × 3=4×3=12. Finding that 12 is contained 6 times in 77, we try 6, as follows: annexing 6 to the treble of the first root-figure, and multiplying the result (66) by 6, we obtain 396, to which we add the trial-divisor (12)— removed two places to the left; multiplying the sum (1596)

by 6, we find that the product, 9576, is not contained in 7783; for which reason the second root-figure must be lower than 6. Substituting 5 for 6, and finding that 7625 is contained in 7783, we write 5 as the second root-figure. Subtracting 7625 from 7783, and bringing down the next period (426), we have, as the second dividend, 158426. Conceiving the last two figures (26) of this dividend cut off, we obtain the second trialdividend, 1584. For the second trial-divisor, we treble the square of the known part of the root: 252 × 3=625 × 3=1875. As 1875 is not contained in 1584, we write o as the third rootfigure. Bringing down the next period (589), we have, as the third dividend, 158426589. Conceiving the last two figures (89) of this dividend cut off, we obtain the third trial-dividend, 1584265. For the third trial-divisor, we treble the square of the known part of the root: 2502 × 3=62500 × 3=187500.* Finding that 187500 is contained 8 times, but not 9 times, in 1584265, we try 8 for the next root-figure: annexing 8 to the treble of the known part of the root, and multiplying the result (7508) by 8, we obtain 60064, to which we add the trialdivisor (187500)-removed two places to the left; multiplying the sum (18810064) by 8, we have for product 150480512, which is contained in 158426589. We therefore write 8 as the next root-figure. Subtracting 150480512 from 158426589, and bringing down the last period (234), we have, as the fourth dividend, 7946077234. Conceiving the last two figures (34) of this dividend cut off, we obtain the fourth trial-dividend, 79460772. For the fourth trial-divisor, we treble the square of the known part of the root: 25082 × 3 =6290064 × 3 =18870192. As 18870192 is contained 4 times, but not 5 times, in 79460772, the units' figure of the root cannot be higher than 4, which we try: annexing 4 to the treble of the known part of the root, and multiplying the result (75244) by 4, we obtain 300976, to which we add the trial-divisor (18870192)-removed two places to the left; multiplying the sum (1887320176) by 4, we have for product 7549280704. which is contained in 7946077234. We therefore set down 4 as the units' figure of the root; and as the subtraction of 7549280704 from 7946077234 gives 396796530 for remainder, 25084 is the cube root (not of 15783426589234, but) of the difference between 15783426589234 and 396796530.

NOTE I.-When a trial-divisor is not contained in the corresponding trial-dividend, write a cipher as the next figure of the root; convert the trial-divisor into a new trial-divisor, by

This trial-divisor is most easily obtained when we annex two ciphers to the second trial-divisor (1875). -*

S

annexing two ciphers; convert the dividend into a new dividend, by bringing down the next period; and convert this new dividend into a new trial-dividend by conceiving the last two figures cut off.

Nore 2.--When the discovered" figures of a cube root are greater in number, by 2. than those remaining to be discovered, the latter can be found by division only, as follows: For dividend, bring down the outstanding periods, and cut off twice as many figures as there are root-figures to be determined; for divisor, treble the square of the known part of the root: the resulting quotient -the remainder being disregarded-will be the rest of the root.

Thus, in extracting 12906416731998560897069, we need not work, in the ordinary way, for more than the first fire rootfigures (23456)-knowing that the total number is eight

(23+1). After the finding of the fifth root-figure (6), we have

merely to bring down the outstanding periods-all but the last sir figures; to divide 1302329182560, the number so obtained, by (23456% % 3)1650551808; and, disregarding the remainder, to write 789--the resulting quotient as the rest of the root:

234562x3=550183936 × 3=1650551808;

1302329182560÷ 1650551808789; required root 23456789.

The reason is this: Putting x for the last three figures of the root, we have 12906416731998560897069=23456000+ x; 12906416731998560897069 = (23456000+x)3=234560003 + 3 x 234560002 x +3 × 23456000 × x2+x3. Subtracting 234560003, therefore, from 12906416731998560897069, we obtain 1302329182560897069=3× 234560002 × x+3×23456000 1302329182560897069. 3x23456000 x x2 3X234560002 As the num

3× 234560002

x3 3× 234560002

=x+

x2

+

=x+
x3

23456000 3X 234560002

+

bers represented by r occupies only three places, that represented by x2 occupies not more than six places (§ 230), and is therefore less than 1,000,000; so that

1,000,000

x2

is less than

1,000,000

But

is less than

23456000
1,000,000

: because

23,456,000 23,456,000 occupies eight places, and the smallest (whole) number which occupies eight places is 10,000,000. Conseis less than 1,000,000

quently,

10,000,000

I

10

x2 or than Again: 23456000 the number represented by 3 occupies not more than nine places (§ 237), and is therefore less than 1,000,000,000; so that

23456000 occupies at least fifteen places (§ 230), and the smallest (whole) number which occupies fifteen places is

100,000,000,000,000. Consequently,

1,000,000,000 100,000,000,000,000'

x2 23456000 3X234560002 than unity. So that the division of 1302329182560897069by 3X 234560002—or of 1302329182560897069 by 1650551808000000, or of 1302329182560 by 1650551808-gives x, the remainder being disregarded.

239. As the cube of a fraction is obtained when the cube of the numerator is divided by the cube of the denominator-so, the cube root of a fraction is obtained when the cube root of the numerator is divided by the cube root of the denominator.