+ 189 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 3 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 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 + 3?. 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 +- 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 2302. 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 ofien it is contained in the dividend, and write the result as the next figure of the root. Add to the trial divisor three times the prodluct of the tens figure of the root by the units 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 dwisor , and proceed as before until all the periods are brought down. Or, RULE 2. — Ilaring 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. Msultiply 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 in Notes 1, 2, 3, and 5, under the rule for the extraction of the square root (Art. 525), are equally applicable to the exo traction of the cube root, except that two ciphers must be placed at the right of a true 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 3 ? 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 1šit? 11. What is the value of 3:311 ? Ans. : 12. What is the value expressed by 34965783 ? Ans. 327. 13. What is the value of 122615327232} ? Ans. 4968. 14. What is the value of 436036824287+? Ans. 7583. Ans. 17. 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+: 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 roots 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. EXAMPLES. 1. Required the fourth root of 50625 ? Ans. 15. BY FACTORS. BY SUCCESSIVE EXTRACTIONS. 5|50625 50625 225 4 510125 42 106 84 225/15, Ans. 1 327 25 125 125 15, Ans. Ans. 999. 3. What is the cube root of 262144 ? Ans. 64. 4. What is the fourth root of 43046721 ? 5. What is the fifth root of 14348907 ? Ans. 27. 6. What is the sixth root of 11390625 ? Ans. 15. 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. OPERATION. Ex. 1. Required the cube root of 92959677. Ans. 453. The greatest cube 0 0 92959677 453 contained in the left16 64 hand period we find to be 64, whose root, 4, 16 28959 we write as the first 32 27125 figure of the required root. This figure, 4, 8 4800 1834677 we write under the ci4 625 1834677 pher of the first column; 120 5425 0 and adding it to the cipher obtain 4, which 5 650 sum multiplied by the 4 gives 16; and 125 607500 the result, 16, we write under the cipher 4059 of the second 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 sub1350 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 1353 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, 120, 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 in the last column bringing down the next period, 677, obtaining 1834677, the work is completed preparatory to finding the third root figure. To determine that figure, as before, we divide the last term of the third column by the last term of the second. We thus obtain 3, which, added to the last term of the first column, gives 1353, which sum, multiplied by the 3, gives 4059; and that product, being adoled to the last term of the second column, gives 611559 ; and that sum, multiplied by the 3, gives 1834677, which being exactly as large as the last term of the third column, on being written under it and subtracted there is no remainder. The given number is therefore a perfect power, and the cube root sought is 453. In practice, the work may be performed with less figures, by writing down in the several columns only the results. RULE. — Commence as many columns as there are units in the exponent of the root to be extracted, by writing the given number as the head of the right-hand column, and a cipher as the head of each of the others. Separate the given number into as many periods as possible of as many figures each as the exponent of the root requires ; and having found the nearest root of the lefi-hand period, write it as the first figure of the required root. Write this figure in the first column, and, having added it to what stands above it, multiply the sum by the same figure, and write the product in the second column ; add, in like manner, in the second column, and multiply the sum by the same figure, writing the product in the third column; and so proceed, writing the last product in the last column, and subtracting it from what stands above it. Then add the same figure to the last term of the first column, multiply the sum by the same figure, and aıld the product to the last term of the second column; and so on, writing the last product in the last column but one. Repeat the process, stopping each time with one column farther to the left, till the last product shall fall in the second column. Add the figure found for the rout to the last term of the first column; annex one cipher to the last number in the first column, two ciphers to the last number in the second column, and so on; and to the last number in the last column bring down the next period for a dividend. Take the last term of the column next to the last for a trial divisor, and see how often it is contained in the dividend, and write the result as the next figure of the root. Add this figure to the last term of the first column, multiplying the sum by the same figure, add the product to the second column, and so on; proceed as before, till all the periods have been brought down, or an answer sufficiently exact has been obtained. Note 1. — When any dividend will not contain its corresponding trial divisor, write a cipher in the root, bring down to the dividend another period. annex an additional cipher to the last term of the first column, two additional |