Since the entire root is to be the side of a cube, let us form a block, (Fig. 1), by the addition of 70494 solid inches, in such a manner as to preserve the cubical form, its size will be that of the required block. To preserve the cubical form, the addition must be made upon three adjacent sides or faces. The addition will therefore be composed of 3 flat blocks to cover the 3 faces, (Fig. II); 3 oblong blocks to fill the vacancies at the edges, (Fig. III); and 1 small cubical block to fill the vacancy at the corner, (Fig. IV). Now, the thickness of this enlargement will be the additional length of the side of the cube, and, consequently, the second figure in the root. To find thickness, we may divide solid Fig. II. contents by surface, or area. But the area of the 3 oblong blocks and little cube cannot be found till the thickness of the addition be determined, because their common breadth is equal to this thickness. We will therefore find the area of the three flat blocks, which is sufficiently near the whole area to be used as a trial divisor. As these are each equal in length and breadth to the side of the cube whose faces they cover, the whole area of the three is 70 X 70 X 3=14700 square inches. This number is obtained in the operation by annexing 2 ciphers to three times the square of 7; the result being written at the left hand of the dividend. Dividing, we obtain 4, the probable thickness of the addition, and second figure of the root. With this assumed figure, we will complete our divisor by adding the area of the 4 blocks, before undetermined. The 3 oblong blocks are each 70 inches long; and the little cube, being equal in each of its dimensions to the thickness of the addition, must be 4 inches long. Hence, their united length is 70 +70 70+4=214. This number is obtained in the operation by multiplying the 7 by 3, and annexing the 4 to the product, the result being written in column I, on the next line below the trial divisor. Multiplying 214, the length, by 4, the common width, we obtain 856, the area of the four blocks, which added to 14700, the trial divisor, makes 15556, the complete divisor; and multiplying this by 4, the second figure in the root, and subtracting the product from the dividend, we obtain a remainder of 3270 solid inches. With this remainder, for the same reason as before, we must proceed to make a new enlargement. But since we have already two figures in the root, answering to the two periods of the given number, the next figure of the root must be a decimal; and we therefore annex to the remainder a period of three decimal ciphers, making 8270.000 for a new dividend. The trial divisor to obtain the thickness of this second enlarge ment, or the next figure of the root, will be the area of three new flat blocks to cover the three sides of the cube already formed; and this surface, (Fig. IV,) is composed of 1 face of each of the flat blocks already used, 2 faces of each of the oblong blocks, and 3 faces of the little cube. But we have in the complete divisor, 15556, 1 face of each of the flat blocks, oblong blocks, and little cube; and in the correction of the trial divisor, 856, 1 face of each of the oblong blocks and of the little cube; and in the square of the last root figure, 16, a third face of the little cube. Hence, 16 +856 +1555616428, the significant figures of the new trial thus: 16, 6, and 6 are 28, and we write the unit figure in the new trial divisor; then 2 to carry, and 5 and 5 are 12, &c. We annex 2 ciphers to this trial divisor, as to the former, and dividing, 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. From this example and analysis 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, 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. 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 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. EXAMPLES FOR PRACTICE. 1. What is the cube root of 79.112? 2. What is the cube root of 84604519? 3. What is the cube root of 2357947691 ? Ans. 439 Ans. 1331, Ans. 22215 4. What is the cube root of 10963240788375? 6. What is the cube root of .091125? Ans. 646866. 7. What is the cube root of .000529475129 ? Ans. .0809 8. What is the approximate cube root of .008649? Ans. .2052+. Extract the cube roots of the following numbers :— APPLICATIONS IN CUBE ROOT. 1. What is the length of one side of a cistern of cubical form, containing 1331 solid feet? Ans. 11 feet. 2. 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. 3. A cubical box contains 474552 solid inches; what is the area of one of its sides? Ans. 421 sq. ft. 4. How much paper will be required to make a cubical box which shall contain of a solid foot? Ans. of a yard. 5. 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. NOTE. Spheres are to each other as the cubes of their diameters or circumferences. 6. There are two spheres whose solid contents are to each other as 27 to 343; what is the ratio of their diameters? ANALYSIS. Since spheres are to each other as the cubes of their diameters, the diameters will be to each other as the cube roots of the spheres ; and 3/27 3,3/3437; hence the diameters required are as 3 to 7. |