as many of the left-hand periods, as there were figures already found in the root. But, regard being paid to the local value of the figures, the process of extracting the root may be considerably shortened. To show the man. ner in which this is effected, we shall again extract t root of 14706125. 882125 0. 360.3(a+b) c. 25 c2. 176425=3(a+b)2+3(a+b) c+c2. 5= c. = ·3 (a+b)2c+3 (a+b) c2 + c3. The process is the same as in the preceding Article, until we have found the second figure of the root; except that we bring down the whole of the second period, at the right of the first remainder, separating the two right-hand figures by an accent. Previously to finding b, the tens of the root, we subtracted a3 from the first period, and our remainder with the succeeding period annexed, 6706, contains b (3 a2+3 ab+b2), together with the thousands arising from the rest of the formula. Having found b= 4 (tens), we are to ascertain the value of 6 (3a2+3 ab +62), and subtract it from the dividend, including the two figures cut off. root. Our divisor is 3 a2=12,(120000), and 3 ab=3.2.4, or rather 3.200.40=24000, is three times the product of the figure last found by the preceding figure of the But since b is of the order of units next below a, the number corresponding to 3ab will contain a significant figure one degree lower than is found in 3 a2. Therefore, 24,3 ab, is to be put under 12, 3 a2, one place farther to the right. We next find b2, = (4)2 = 16, or rather (40)2=1600, and since it contains a significant figure one degree lower than 3 ab, we put 16 under 24, = 3 ab, one place farther to the right than this last. Adding these three numbers, as the figures now stand, we have 1456 (hundreds), = 3 a2+3 ab+b2, which we multiply by 4 (tens), = b, and obtain 5824 (thousands),= 3a2b+3a b2+b3. We subtract this product, 5824, from the dividend, including the two figures rejected in the division, bring down the next period to the right of the remainder, and have 882125=3(a+b)2c+3 (a+b) c2 +c3=[3 (a+b)2+3 (a+b) c + c2] c. Rejecting the two right-hand figures of 882125, we take the rest as a dividend, and divide by 1728 (hundreds), = 3(a+b)2, that is, by three times the second power of the hundreds and tens of the root. The division gives 5, = c, which we place as the units of the root. We now wish to find the value of [3 (a+b)2 + 3 (a + b) c + c2] c, and subtract it from 882125. Our divisor is 3 (a+b)2,: = 1728 (hundreds), and 3(a+b) c = 3 .24 . 5 360, or rather 3.240.5: = 3600, is the product of three times the figure last found by the preceding figures of the root; and as this product would, if the last figure of it did not happen to be zero, contain a significant figure one degree below the value of 3 (a+b), we put 360 under 1728, one place farther to the right. We now put 25, = c2, which contains a significant figure still one degree lower, under 360, one place still farther to the right. Adding these numbers, as the figures now stand, we have 1764253 (a + b)2 + 3 (a+b) c + c2, which we multiply by 5,= c, and have 882125 = 3 (a + b)2 c + 3(a+b) c2 + c3. This subtracted from the last dividend, including the rejected figures, leaves no remainder. Hence, the work is complete. ART. 99. From the preceding analysis we deduce the following RULE FOR EXTRACTING THE THIRD ROOTS OF NUMBERS. 1. Commencing at the right, separate the number, by means of accents, into periods of three figures each; the left-hand period may contain one, two, or three figures. 2. Find the greatest third power in the left-hand period, place its root at the right, and subtract the power from that period. 3. To the right of the remainder bring down the next period, separating the two right-hand figures by an accent; those to the left of the accent will form a dividend. For a divisor take three times the second power of that part of the root already found. Divide the dividend by the divisor, and put the quotient as the second figure of the root. 4. Take three times the product of the figure last found by the preceding part of the root, and place it under the divisor, one place farther to the right; under which, one place farther to the right, place the second power of the figure of the root last found. Add together the divisor and the numbers placed under it, as the figures stand, and multiply the sum by the figure of the root last found. Subtract this product from the dividend, including the two rejected figures. 5. To the right of the remainder bring down the next period, forming a new dividend, in the same manner as the first was formed. Take for a divisor three times the second power of the whole root so far as found; divide and place the quotient as the next figure of the root. 6. Find three times the product of the last figure by the whole of the preceding part of the root, and put it under the divisor, one place farther to the right; under this, one place farther to the right, put the second power of the last figure of the root found. Add the divisor and the numbers placed under it, as the figures stand, multiply the sum by the last figure of the root found, and subtract the product from the dividend, including the rejected figures. 7. Repeat the operations stated in the 5th and 6th parts of the rule, until the given number is exhausted. Remark 1st. Whenever the divisor is not contained in the dividend, or the figures to the left of the two rejected, put a zero in the root, and bring down the next period, separating the two right-hand figures; the divisor for finding the next figure of the root will then be the same as before, except with the annexation of two zeros. Remark 2d. Whenever the number to be subtracted exceeds that from which it is to be taken, diminish the last figure found in the root, until a number is obtained which can be subtracted. 1. What is the third root of 525557943 ? |