We annex 2 ciphers to this trial divisor, as to the former. Or we may obtain this trial divisor, in the same way as the first one, by taking 3 times the square of 74 (which we regard as the first part), with one cipher annexed. Dividing, we obtain 5, the third figure in the root. To complete the second trial divisor, after the manner of the first, the correction may be found by annexing .5 to 3 times the former figures, 74, and multiplying this number by .5. But as we have, in column I, 3 times 7, with 4 annexed, or 214, we need only multiply the last figure, 4, by 3, and annex .5, making 222.5, which multiplied by .5, gives 111.25, the correction required. Then we obtain the complete divisor, 16539.25, the product, 8269.625, and the remainder, .375, in the manner shown by the former steps. The above is the geometrical explanation. The following is another explanation of the same process : We find that the greatest cube in 413494 is 343000, whose cube root is 70, which we write at the right. Since the cube of a number divided into any two parts is equal to the cube of the first part, plus 3 times the square of the first by the second, plus 3 times the first by the square of the second, plus the cube of the second part (424), therefore, having found the cube of the first part 70, which is 343000, the remainder 70494 must be equal to 3 times the square of the first part by the second part (?), plus 3 times the first part by the square of the second part, plus the cube of the second part. Since we do not know the second part, we take as a trial divisor 3 times the square of the first, which is 14700, and we find that the second part is about 4. The product of the quotient by the trial divisor will be 3 times the square of the first by the second. To this must be added 3 times the first by the square of the second = 3 x 70 x 42 (but since the 4 in the quotient forms one of these factors we add to the trial divisor 3 x 70 x 4 = 840), and the cube of the second, making the square 4 x 4 for addition to the trial divisor, and our complete divisor will be 14700 + 840 + 16=15556, which multiplied by the quotient figure 4 gives a product of 62224. With the remainder, we proceed as before and the root is 74.5+. 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 lefthand 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, and 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 I 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. VIII. If there is a remainder after the root of the last period is found, annex periods of ciphers and proceed as before. The figures of the root thus obtained will be decimals. 1. If at any time the product is greater than the dividend, diminish the trial figure of the root, and correct the erroneous work. 2. If a cipher occurs in the root, 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. The cube root of a fraction may be found by extracting the cube root of the numerator and denominator. In extracting the cube root of decimal numbers, begin at units' place and proceed both toward the left and the right, to separate into periods of three figures each. 122 10088 244 5044 529200 5024000 1269 11421 540621 4865589 55212300 158411000 12872 25744 55238044 110476088 5526379200 47934912000 128768 1030144 5527409344 44219274752 3715637248, Rem. Ans. 439. 3. What is the cube root of 84604519? 4. What is the cube root of 2357947691? Ans. 1331. 5. What is the cube root of 10963240788375? Ans. 22215. Ans. 646866. 6. What is the cube root of 270671777032189896 ? 7. What is the cube root of .091125? 8. What is the cube root of .000529475129 ? Ans. .45. Ans. .0809. 9. What is the approximate cube root of .008649? Ans. .2052+. Extract the cube roots of the following numbers: 10. 2. Ans. 1.259921+. 13. 5. 11. 3. Ans. 1.442249+. 14. 6. 12. 4. Ans. 1.587401+. 15. 7. Ans. 1.709975+. Ans. 1.817120+. APPLICATIONS OF CUBE ROOT. 441. The following principles of geometry afford applications in cube root. PRINCIPLES. — I. The ratio of two similar solids is equal to the cube of the ratio of any two like dimensions. II. The ratio of' any two like dimensions of similar solids is equal to the cube root of the ratios of the solids. EXAMPLES. a 1. The lengths of two similar solids are 4 in. and 50 in. The first contains 16 cu. in. What does the second contain ? Ans. 31250 cu. in. 2. If a ball 5 in. in diameter weighs 8 pounds, what will be the weight of a similar ball 10 in. in diameter ? 3. I have two cubical boxes. The side of the smaller contains 6 in., and of the larger, 10 in. How many times would the smaller contain the larger ? 4. What is the length of one side of a cistern of cubical form, containing 1331 solid feet ? Ans. 11 feet. 5. The pedestal of a certain monument is a square block of granite, containing 373248 solid inches. What is the length of one of its sides ? Ans. 6 feet. 6. A cubical box contains 474552 solid inches. What is the area of one of its sides ? Ans. 424 sq. ft. 7. A man wishes to make a bin to contain 125 bushels, of equal width and depth, and length double the width. What must be its dimensions ? Ans. Width and depth, 51.223+ inches; length, 102.446+ inches. 8. If a cable 4 in. in circum. will support a sphere 2 ft. in diam., what is the diam. of that sphere which will be supported by a capie 5 in. in circum.? Ans. 2.32+ ft. а ROOTS OF ANY DEGREE. 442. Any root whatever may be extracted by means of the square and cubic roots, as will be seen in the two cases which follow. EXAMPLES. 443. When the index of the required root contains no other factor than 2 or 3. We have seen that if we raise any power of a given number to any required power, the result will be the power of the given number denoted by the product of the two indices (419). Conversely, if we extract successively two or more roots of a given number, the result must be that root of the given number denoted by the product of the indices. 1. What is the 6th root of 2176782336 ? SOLUTION. The index of the required root is 6 = 2 × 3; we therefore extract the square root of the given number, and the cube root of this result, and obtain 36, which must be the 6th root required. Or, we first find the cube' root of the given number, and then the square root of the result, as in the operation. Or, 3 2176782336 = 1296. 1296 RULE. 36, Ans. Separate the index of the required root into its prime factors, and extract successively the roots indicated by the several factors obtained; the final result will be the required root. Ans. 43. 2. What is the 6th root of 6321363049 ? Ans. 32. Ans. 543. |