will be denoted by (a + b) = a + 3 a' b + 3 ab + b. Then, by this formula, if a = 3, and b equal 6, we have 3 tens + 6 units 30 +6 = 36, and 363 (30 + 6) = 303 + 3 (302 X 6) + 3 (30 X 6') + 63 = 46656. Or, analytically, a + 6 = 30 + 6 36 a +6 = 30 + 6 36 (a+b) Xa= 30° + 30 x 6 1080 (a+b) Xb= 30 X 6+ 62 216 (a + b)2 = 303 + 2 X 30 X 6+ 62 1296 a+ = 30 + 6 36 (a + b)2 Xa= 303 + 2 X 302 X 6+30 X 62 - 38880 (a+b)xb= 302 X 6+2 X 30 X 6?+69= 7776 (a+b)3 = 30° +3 (30X 6) +3 (30 x 6?)+6= 46656 It is evident, as evolution is the reverse of involution, that from this process of obtaining a cube may be deduced a method of extracting the cube root. Since the cube of a + b is a+ 3 ab + 3 a b + bo, the cube root of a + 3 a2 b + 3 a b + b must be a + b. Now a, the first term of the cube root, is the cube root of a, the first term of the cube; and if a be subtracted, there will remain 3 a' b + 3 a b? + b*, from which b, the second term of the root, is to be obtained. But 3 a’b + 3 a b? + 63 is the same as (3 a’ + 3 ab + b?) Xb; therefore the remainder equals (3 a2 + 3 ab + b) X b. But as 3 ab, three times the tens into the units, plus b', the square of the units, is generally much less than 3 a’, three times the square of the tens, we consider that 3 a’ x b is about equal to the whole remainder, and taking 3 a' (which we know) as the trial divisor, we obtain b, the units. But as the true divisor is 3 a’ + 3 ab + b2 we add three times the tens by the units plus the square of the units, and multiply the sum by the units, which gives a product equal the whole remainder, or 3 a' b + 3 a b + 69. Since every number of more than one figure may be considered as composed of tens and units, we may have tens and units of units, tens and units of tens, tens and units of hundreds, &c. Hence, the principle just explained applies equally whether the root contains two or more than two figures. 529. To extract the cube or third root of numbers. OPERATION 33 over PROOF. Ex. 1. Beginning at 46656|36 the right, we sep 27 arate the given Trial div., 3 x 302 2700 number into pe 19656 3 X 30 X 6 540 riods, by placing 62 36 a point over the units figure and True divisor, 3276 x 6 19656 the third figure to the left. 0 Since the number of periods is two, 36 x 36 x 36 = 46656 the root will con sist of two figures, tens and units. Then 46656 the cube of 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. The cube of tens is thousands, and must therefore be found in the thousands of the number. The greatest number of tens whose cube does not exceed 46 thousands is 3, which we write as the tens figure of the root. We then subtract the 27 thousands, the cube of the 3 tens, from the 46 thousands, and there remain 19 thousands; and, annexing the next period, we have as the entire remainder, 19656, 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 the product of three times the square of the tens, plus three times the tens into the units, plus the square of the units, multiplied by the units. By dividing this remainder by three times the square of the tens of the root, we obtain the units, or a number somewhat too large. Although it may be too large, it cannot be too small, since the remainder 19656 contains not only three times the square of the tens into the units, but three times the tens into the square of the units, plus the cube of the units. We therefore make three times the square of the tens of the root, 27 hundreds, a trial divisor, with which we divide the 196 hundreds of the remainder, disregarding the 56 units, since they cannot form any part of the product of the square of the tens by the units. The quotient figure obtained, 7, must be the units figure of the root, or a number somewhat larger. But on undertaking to complete the divisor on the supposition that is the true units figure of the root, we find a divisor too large for the remainder. We therefore take 6, a number one less, and to determine whether it expresses the real number of units in the root, we add to the 27 hundreds of the trial divisor three times the 3 tens of the root into the 6 units, plus the square of the 6 units; and multiplying the true divisor, 3276, thus formed, by the units, and subtracting the product, 19656, from the remainder, there is nothing left. Ilence, 46656 is a perfect cube, and 36 its cube root. 2. What is the cube root of 12326391 ? Ans. 231. FIRST OPERATION. = 32 = 1 2 3 2 6 3 9 12 31 23 8 Trial divisor, 3 X 202 -= 1200 4 3 2 6 3 X 20 X 3 180 9 True divisor, 1389 X 3 4167 Trial divisor, 3 x 2302 158700 1 5 9 3 91 1 True divisor, 159391 X 1 15 9 3 9 1 Since there are 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)', 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. 1 2 3 2 6 39 i 231 23 8 Trial divisor, 1200 4 3 2 6 189 41 67 15 9 39 1 691 SECOND OPERATION. + 189 In the second operation the work is somewhat abridged by render ing 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 + 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 + 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 prorluct annex two ciphers 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 to the trial divisor three times the product 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 divisor ; 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 faund, 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 remaindler 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 triul dirisor ; 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 ex 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 jaz? Ans. 4. 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 111? 11. What is the value of 9319 Ans. 11. 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 requireil, 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. |