1st. That a root consisting of 1 place may have from 1 to 3 places in the cube. 2d. That in all cases the addition of 1 place to the root adds 3 places to the cube Hence, I. If we point off a number into three-figure periods, commencing at the right hand, the number of full periods and the left hand full or partial period will indicate the number of places in the cube root. To ascertain the relations of the several figures of the root to the periods of the number, observe that if any number, as 5423, be decomposed, the cubes of the parts will be related in local value, as follows: 50003 54003 125 000 000 000 157 364 000 000 II. The cube of the first figure of the root is contained wholly in the first period of the power; the cube of the first two figures of the root is contained wholly in the first two periods of the power; and so on To learn the combinations of tens and units in the formation of a cube, take any number consisting of two figures, as 54, and decompose it into two parts, 50+4; then having formed the square by 656, III, multiply each part of this square by the units and tens of 54 separately, thus, 542= 502 + 2 x 50 × 4+ 42 502 × 4+2 × 50 × 42 + 43 50+ 2 x 50 x 42 502 × 4+ 3 x 50 x 42+ 43 156924 543- = Of these combinations, the first is the cube of 50, the second is 3 times the square of 50 multiplied by 4, the third is 3 times 50 multiplied by the square of 4, and the fourth is the cube of 4. Hence, III. The cube of a number composed of tens and units is equal' to the cube of the tens, plus three times the square of the tens multiplied by the units, plus three times the tens multiplied by the square of the units, plus the cube of the units. By observing the manner in which the cube is formed, we perceive that each of the last three parts contains the units as a factor; these 1 parts, considered as one number, may therefore be separated into two factors, thus, (3 × 502 + 3 × 50 × 4+ 42) × 4. Hence, IV. If the cube of the tens be subtracted from the entire cube, the remainder will be composed of two factors, one of which will be three times the square of the tens plus three times the tens multiplied by the units plus the square of the units; and the other, the units 1. What is the cube root of 145780726447 ? ANALYSIS. Pointing off the given number into periods of 3 figures each, the four periods show that there will be four figures in the root, (I). Since the cube of the first figure of the root is contained wholly in the first period of the power, (II), we seek the greatest cube in the first period, 145, which we find by trial to be 125, and we place its root, 5, for the first figure of the required root, and regard it as tens of the next inferior order, (654). We now subtract 125, the cube of this figure, from the first period, 145, and bringing down the next period, obtain 20780 for a remainder. And since the cube of the first two figures of the root is contained wholly in the first two periods of the power, (Ii), the remainder, 20780, must contain at least the product of the two factors, one of which is three times the square of the first figure (tens), plus three times the first figure multiplied by the second (units), plus the square of the second; and the other, the second figure (IV) Now if we could divide this remainder by the first of these factors, the quotient would be the other factor, or the second figure of the root. But as the first factor is composed in part of the second figure, which we have not yet found, we can not now obtain the complete divisor; and we therefore write three times the square of the first figure, regarded as tens, or 502 × 3 = 7500, at the left of the dividend, for a trial divisor. Dividing the dividend by the trial divisor, we obtain 2 for the second, or trial figure of the root. To complete the divisor, we must add to the trial divisor, as a correction, three times the tens of the root already found multiplied by the units, plus the square of the units, (IV). But as 50 × 3 × 2 + 22 = (50 × 3+2) × 2, we annex the second figure, 2, to three times the first figure, 5, and thus obtain 50 × 3 + 2 = 152, the first factor of the correction, which we write in the column marked I. Multiplying this result by the 2, we have 304, the correction, which we write in the column marked II. Adding the correction to the trial divisor, we obtain 7804, the complete divisor. Multiplying the complete divisor by the trial figure of the root, subtracting the product from the dividend, and bringing down the next period, we have 20780 for a remainder. We have now taken the cube of the first two figures of the root considered as tens of the next inferior order, from the first three periods of the number; and since the cube of the first three figures of the root is contained wholly in the first three periods of the power, (II), the remainder, 20780, must contain at least the product of the two factors, one of which is three times the square of the first two figures of the root (regarded as tens of the next order) plus three times the first two figures multiplied by the third, plus the square of the third; and the other, the third figure, (IV). Therefore, to obtain the third figure, we must use for a trial divisor three times the square of the first two figures, 52, considered as tens. And we observe that the significant part of this new trial divisor may be obtained by adding the last complete divisor, the last correction, and the square of the last figure of the root, thus: This number is obtained in the operation without re-writing the parts, by adding the square of the second root figure mentally, and combining units of like order, thus: 4, 4, and 4 are 12, and we write the unit figure, 2, in the new trial divisor; then 1 to carry and 0 is 1; then 3 and 8 are 11, etc. Annexing two ciphers to the 8112, because 52 is regarded as tens of the next order, and dividing by this new trial divisor, 811200, we obtain 6, the third figure in the root. To complete the second trial divisor, after the manner of completing the first, we should annex the third figure of the root, 6, to three times the former figures, 52, for the first factor of the correction. But as we have in column I three times 5 with the 2 annexed, or 152, we need only multiply the last figure, 2, by 3, and annex the third figure of the root, 6, which gives 1566, the first factor of the correction sought, or the second term in column I. Multiplying this number by the 6, we obtain 9396, the correction sought; adding the correction to the trial divisor, we have 820596, the complete divisor; multiplying the complete divisor by the 6, subtracting the product from the dividend, and bringing down the next period, we have 249150947 for a new dividend We may now regard the first three figures of the root, 526, as tens of the next inferior order, and proceed as before till the entire root, 5263, is extracted. 661. From these principles and illustrations we deduce the following RULE. I. Point off the given number into periods of three figures each, counting from units place toward the left and right. II. Find the greatest cube that does not exceed the left hand period, and write its root for the first figure in the required root; subtract the cube from the left hand period, and to the remainder bring down the next period for a dividend. III. At the left of the dividend write three times the square of the first figure of the root, and annex two ciphers, for a trial divisor; divide the dividend by the trial divisor, and write the quotient for a trial figure in the root. IV. Annex the trial figure to three times the former figure, and write the result in a column marked I, one line below the trial divisor, multiply this term by the trial figure, and write the product on the same line in a column marked II; add this term as a correction to the trial divisor, and the result will be the complete divisor. V. Multiply the complete divisor by the trial figure; subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. VI. Add the square of the last figure of the root, the last term in column II, and the complete divisor together, and annex two ciphers, for a new trial divisor; with which obtain another trial figure in the root. VII. Multiply the unit figure of the last term in column 1 by 3, and annex the trial figure of the root for the next term of column I; multiply this result by the trial figure of the root for the next term of column II; add this term to the trial divisor for a complete divisor, with which proceed as before. NOTES.-1. If at any time the product be greater than the dividend, diminish the trial figure of the root, and correct the erroneous work. 2. If a cipher occur in the roct, annex two more ciphers to the trial divisor, and another period to the dividend; then proceed as before with column I, annexing both cipher and trial figure. EXAMPLES FOR PRACTICE. 1. What is the cube root of 389017? 7. What is the cube root of .000148877? Ans. 73. Ans. 354. Ans. 2187. Ans. 4624. Ans. 3.12866 +. Ans. 4.8076 + Ans. .053. Ans. 4968. Ans. 8. Ans. 1156. 12. How much does the sum of the cube roots of 50 and 31 exceed the cube root of their sum? CONTRACTED METHOD. Ans. 2.4986 +. 662. In applying contracted decimal division to the extraction of the cube root of numbers, we observe, 1st. For each new figure in the root, the terms in the operation extend to the right 3 places in the column of dividends, 2 places in the column of divisors, and 1 place in column I. Hence, 2d. If at any point in the operation we omit to bring down new periods in the dividend, we must shorten each succeeding divisor 1 place, and each succeeding term in column I, 2 places. 1. What is the cube root of 189, correct to 8 decimal places? |