Since there am three periods the root will contain three figures, the first two of which may be considered as tens and units of tens. Aj the cube of tens cannot give less than thousands, we must find that cube in the two left-hand periods; and as we have tens and units of tens, their cube will equal the cube of the tens, plus three times the square of the tens into the units, plus three times the tens into the square of the units, plus the cube of the units. This we apply to the first two periods exactly the same as when the root consists of but two figures, and thus take from the given number the cube of the 23 tens, which leaves a remainder of 159391. We now consider the given number, 159391, as the cube of a number consisting of 23 tens and a certain number of units, which cube will of course equal the cube of the tens, plus three times the square of the tens into the units, plus three times the tens into the square of the units, plus the cube of the units. But the cube of the tens, or (23)3, has already been taken from the given number, leaving a remainder, 159391, which must equal three times the square of the tens into the units, plus three times the tens into the square of the units, plus the cube of the units; or equal the product of three times the square of 23, plus three times 23 into the units, plus the square of the units, multiplied by the units. From this we readily obtain the units, just as when we had but two figures in the required root. In the second operation the work is somewhat abridged by rendering shorter the method of finding each divisor after the first, of both kinds. Thus, the second true divisor is obtained by prefixing to the second root figure, 3, three times the part of the root preceding it, or 8 X 2 = 6, and adding 189, the product of the number 63 thus formed by the last root figure, 3, to the preceding trial divisor, or 1200 -4- 189 = 1389, second true divisor. This is equivalent to adding to the trial divisor, in forming the true divisor, three times the tens into the units, plus the square of the units, or 3 X 20 X 3 -f 32. For the second trial divisor we annex two ciphers to the sum found by adding together the square of the last root figure, the last true divisor, and the number standing over it, or 9 -)- 1389 -f- 189 = 1587, with two ciphers annexed = 158700, the second trial divisor. This is equivalent to taking for the trial divisor three times the square of the tens, or the part of the root already found, or 3 X 230*. The next true divisor is found in like manner as was the second true divisor. Rule 1. — Separate the given number into as many periods as possible of three figures each, beginning at the units' place. Find the greatest cube in the left-hand period, and write its root as the first figure of the required root. From that period subtract the cube, and to the remainder bring down the next period for a dividend. Multiply the square of the root figure by 3, and to the product annex two ciphers for a trial divisor, and see how often it is contained in the dividend, and ivrite the result as the next figure of the root. Add to the trial divisor three times the product of the tens figure of the root by the un its figure with a cipher annexed, and the square of the last figure, for a true divisor Multiply the true divisor by the last figure of the root; subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. Multiply the square of the root figures already found by 3, and to the product annex two ciphers for a new trial diviior; and proceed as before until all the periods are brought down. Or, Rule 2. —Having found the first trial divisor and determined the second root figure as by the preceding rule, — Take three times the part of the root already found, except the last figure, to it annex the last figure of the root, multiply the result by the figure annexed, and write the product below the trial divisor, and add it to the same for a true divisor. Multiply the true divisor by the last figure of the root: subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. To the last true divisor and the number immediately over it, add the square of the last root figure, and to the sum annex two ciphers for a new trial divisor; then proceed as before. Note 1. — The observations made hi Notes 1, 2, 3, and 5, under the rale fof the extraction of the square root (Art. 625), are equally applicable to the extraction of the cube root, except that two ciphers must be placed at the right of a tme divisor when it is not contained in its corresponding dividend, and in pointing off decimals each period must contain three figures. Note 2. — If the given number is a common fraction, reduce it to its simplest form, if it is not so already, and extract the root of both terms, if they be perfect powers; otherwise, either find the product of the numerator by the square of the denominator, extract its root, and divide the result by the denominator; or reduce the fraction to a decimal, and extract the root of the decimal. Examples. 3. What is the cube root of 77308776? Ans. 426. 4. What is the cube root of J? Ans. \. 5. What is the cube root of 84.604519? Ans. 4.39. 6. What is the cube root of 54439939? Ans. 379. 7. What is the cube root of 60236288? Ans.. 392. 8. Required the cube root of .726572699. Ans. .899. 9. Extract the third root of 109215352. Ans. 478. 10. What is the'third root of 4JH? Ans. jf. 11. What is the value of V^tM? Ans- i !• 12. What is the value expressed by \ 31965783? Ans. 327. 13. What is the value of 122615327232*? Ans. 4968. 14. What is the value of 436036824287*? Ans. 7583. 530. When the cube root is to be extracted to many places of decimals, the work may be contracted thus : — Having found in the usual way one more than half of the root figures required, the rest may be found by dividing the last remainder by its corresponding true divisor, as in contracted division of decimals (Art. 276), observing however at each step to reject two figures from the right of the divisor and one from the right of the remainder. Example. 1 Required the cube root of 2 to four places of decimals. Ans. 1.2599-f-. 2. Find the third root of 11 to four places of decimals. Ans. 2.2239. 3. Extract the cube root of 3 to six places of decimals. Ans. 1.442249+. 4. Extract the cube root of 9 to fifteen places of decimals. Ans. 2.08008382301904. EXTRACTION OF ANY ROOT. 531. The root corresponding to any perfect power may be obtained by resolving that power into its prime factors, and multiplying together one of each number of equal factors denoted by the exponent of the required root. Thus one of each two equal factors of the power will give the second or square root; one of each three equal factors will give the third or cube root; one of each four equal factors will give the fourth root; and so on. 532. When the index or exponent of the root to be extracted is a composite number, the root may be obtained by successive extractions of the simpler roofs denoted by the several factors of that exponent. Thus the fourth root may be obtained by extracting the square root twice in succession; the sixth root by extracting the square root and then the cube root; and so on. 533.° When the given number is an imperfect power, or otherwise, or the exponent denoting the root is prime, or otherwise, the required root may be found by an elegant process, perfect in principle, called from its inventor, Horner's Method. Ex. 1. Required the cube root of 92959677. Ans. 453. Operation The greatest cube 0 0 92959677l453 contained in the left 4 l6 0^ i hand period we find to — — be 64, whose root, 4, 4 16 28959 we write as the first 4 32 27125 figure of the required — root. This figure, 4, 8 4800 1804677 we wr|te under the ci 4 625 1834677 pher of the first column; 120 5425 ~0 »pd ad(|\DS * to,th? cipher obtain 4, which ° sum multiplied by the 4 gives 16; and 125 607500 tne resu't, 16, we write under the cipher 5 4059 of tne 8eeond column, and by addition obtain 16, which sum, multiplied by the 4, 130 611559 gives 64; and the result, 64, we write in 5 the last column under the left-hand period, —— 92, of the given number. The 64 sub 1350 tracted from the 92 above it gives for a remainder 28. 3 We next add the 4 to the last term, 4, of the first - „.„ column, obtaining 8; and the result, 8, multiplied by the iooo 4, gives 32, which we write under the last term, 16, of the second column, and, adding the same together, obtain 48. We next add the 4 to the last term, 8, of the first column, obtaining 12; and, annexing one cipher to the last term, 12, of the first column, obtaining 120, two ciphers to the last term of the second column, obtaining 4800, and to the remainder in the last column bringing down the next period, 959, obtaining 28959, we complete the work preparatory to the finding of the second root figure. To determine that root figure, we take the last term, 4800, of the second column, for a trial divisor, and the last term, 28959, of the last column, for a dividend: and, dividing, 6 would appear to be the second figure of the root. This, on trial, however, is found to be too large; we therefore take 5, which answers. This 5 we add to the last term, 1 20, of the first column, obtaining 125; which sum, 125, multiplied by the 5, gives 625, and that product, added to the last term, 4800, of the second column, gives 5425; and this result, 5425, multiplied by the 5, gives 27125, which, written in the last column and subtracted from the figures above it, gives a remainder 1834. Then we add the 5 to the last term, 125, of the first column, obtaining 130; and the result, 130, multiplied by the 5, gives 650, which we write under the last term, 5425, of the second column, and by addition obtain 6075. We next add the 5 to the last term, 130, of the first column, obtaining 135; and, annexing one cipher to the last term, 135, of the first column, obtaining 1350, two ciphers to the last term, 6075, of the second column, obtaining 607500, and to the remainder |