NEW METHOD. 826. The New Method of extracting cube root is shorter and more convenient than the ordinary method. The abbreviation consists in obtaining the true and trial divisors by a law which enables us to use our previous work. NOTE. This method seems to have been approximated by several writers, although I have not found any who present it in the form in which it is here given. 1. Extract the cube root of 14706125. SOLUTION. We find the OPERATION. 14.706.125(245 TEE number of figures in the root, 1ST COL. 1728. . t. d. 176425 C. D. 882125 dividend, and divide it by the number in Col. 2d, as a trial divisor, for the second term of the root. We then take 2 times 2, the first term, and write the product 4 in Col. 1st, under the 2, and add; then annex the second term of the root to the 6 in Col. 1st, making 64, and multiply 64 by 4 for a correction, which we write under the trial divisor; and adding the correction to the trial divisor, we have the complete divisor, 1456. We then multiply the complete divisor by 4, subtract the product from the dividend, and annex the next period for a new dividend. We then square 4, the second figure of the root, write the square under the complete divisor, and add the correction, the complete divisor, and the square for the next trial divisor, which we find to be 1728. Dividing by the trial divisor, we find the next term of the root to be 5. We then take 2 times 4, the second term, write the product 8 under the 64, add it to 64, and annex the third term of the root to the sum, 72, making 725, and then multiply 725 by 5, giving us 3625 for the next correction. We then find the complete divisor, by adding the correction to the trial divisor; multiply the complete divisor by 5, and subtract, and we have no remainder. NOTE. The correctness of this method may readily be seen by using letters and following the changes indicated in the solution. Rule.-I. Separate the number into periods of three figures each; find the greatest number whose cube is contained in the first period, and write it in the root. II. Write the first term of the root at the head of the 1st Col.; 3 times its square, with two dots annexed, at the head of 2d Col., and its cube under the first period; subtract and annex the next period to the remainder for a dividend; divide the number in 2d Column as a TRIAL DIVISOR, and place the quotient as the second term of the root. III. Add twice the first term of the root to the number in the first column; annex the second term of the root, multiply the result by the second term, and write the product under the trial divisor for a CORRECTION; add the CORRECTION to the TRIAL DIVISOR, and the result will be the COMPLETE DIVISOR; multiply the COMPLETE DIVISOR by the last term of the root, subtract the product from the dividend, and annex the next period to the result for a new dividend. IV. Square the last term of the root, and take the sum of this SQUARE, the last COMPLETE DIVISOR, and the last CORRECTION, annexing two dots, for a new TRIAL DIVISOR; divide the dividend by this for the next term of the root. V. Add twice the second term of the root to the last number in the first column; annex the last term of the root to the sum, multiply the result by the last term, and write the product under the last trial divisor for a CORRECTION; add the CORRECTION to the TRIAL DIVISOR, and the result will be the COMPLETE DIVISOR; use this as before, and thus continue until all the periods have been used. NOTE. This rule is indicated in the following formula: 1. COMPLETE DIVISOR TRIAL DIVISOR PRODUCT. 2. TRIAL DIVISOR = PRODUCT+COMPLETE DIVISOR SQUARE. Find the value of the following expressions: 7. 1.72891.125_3. 8. (15625—46656) × ($581)2. 9. 512+÷3375-7x.729. Ans. 37365.1453125. Ans. 662. Ans. 26638. HORNER'S METHOD. 827. Horner's Method is derived from the general method of solving cubic and higher equations invented by Mr. Horner, of Bath, England. 1. Extract the cube root of 14706125. SOLUTION. We write the first term of the root 2, in the 1st col., its square, 4, in 2d col., and its cube, 8, under 1st period, subtract, and bring down the next period. We then add the first term of the root, 2, to 2, the first term in 1st col., multiply the sum 4 by the root, and place it under the 4 in 2d col., take the sum, and the result 12 is our 1st trial divisor; before using it, however, we add 2, the 1st term of the root, to the number 4 in 1st col., giving 6. We then find the 2d term of the root to be 4, annex it to the 6 in the 1st col., multiply the result 64 by 4, the 2d term of the root, place it under the trial divisor, removing it two places to the right, add, and we have the true divisor. We then multi 2 4 OPERATION. 1ST COL. 2d COL. 14.706.125(245 4 8 8 6706 64 12 t. d. 256 5824 725 1456 T. D. 272 1882125 1728 t. d. 3625 68 ply, subtract, and bring down the next period. We then add the 2d term of the root, 4, to the last number in the 1st col., making 68, multiply the result by the last term of the root, 4, write the result under the true divisor, add, and the sum is the next trial divisor, before using which we add the last term of the root, 4, to the last number in 1st col., making 72. We then find the next term of the root, annex it to the last number in 1st col., 72, multiply the result by the last term of the root, write the result removed two places to the right, under the trial divisor, add, and the sum is the true divisor, etc. Rule.-I. Begin at units and separate the number into periods of three figures each, and find the greatest number whose cube is contained in the left hand period. II. Write the first term of the root at the left for the first term of the 1ST COL., and its square for the first term of the 2D COL., and its cube under the left hand period; subtract and annex to the remainder the next period for the FIRST DIVIDEND. III. Add the first term of the root to the first term of 1ST COL., for its second term; multiply the second term by the root found, and add the product to the first term of 2D COL., for a TRIAL DIVISOR, before using which add the root to the last term in 1ST COL. IV. Find the second term of the root by dividing the DIVIDEND by the TRIAL DIVISOR with two ciphers annexed; annex this second term of the root to the last term in 1ST COL., multiply the result by the second term of the root, and add the product advanced two places to the right to the trial divisor, and the result will be the TRUE DIVISOR. V. Multiply the true divisor by the last term of the root found, subtract the result from the dividend, annex the next period to the remainder for the next dividend, and proceed in like manner until all the periods have been used. NOTE.-Require the pupils to apply this method to the problems given under the preceding rule. CONTRACTIONS IN CUBE ROOT. 828. The Rule for Contracted Method is as follows: Rule.-Extract the cube root, as usual, until one more than half the terms required in the root have been found; then with the last divisor and last remainder, proceed as in contracted division to find the other terms of the root, dropping two figures instead of one from the divisor at each step, and one from each remainder. APPLICATIONS OF CUBE ROOT. 829. The Applications of cube root to problems involving geometrical volumes, such as cubes, parallelopipedons, spheres, etc., are extensive. 830. The Edge of a cube is equal to the cube root of its contents. EXAMPLES FOR PRACTICE. 1. Required the dimensions of a cubical cistern which contains 3375 cubic feet. Ans. 15 ft. 2. Required the entire surface of a cubical block which contains 4096 cubic meters. Ans. 1536 sq. meters. 3. Required the edge of a cube equivalent to a solid 40 ft. 8 in. long, 20 ft. 6 in. wide, and 12 ft. 10 in. high. Ans. 22.034 ft. 4. A miller wishes to make a cubical bin which shall contain 100 bu. of grain; what must be its depth? Ans. 4.992 ft. 5. How many square feet of boards will it take to line the four sides of a cubical cistern which contains 300 barrels of water? Ans. 467.42 sq. ft. 6. What would it cost to plaster the bottom and sides of a cubical reservoir which contains 200 barrels of water, at 5 cents a square foot? Ans. $22.29. 7. A farmer wishes to have a bin made whose width shall equal its depth, and length equal 3 times its width, and which shall contain 150 hectoliters of grain; required its dimensions. Ans. Length, 16.83+ ft.; width and depth, 5.61 ft. 8. There is a granary whose capacity is 5000 bushels; its length is twice its breadth, and breadth twice its height; required its dimensions. Ans. 36.784 ft.; 18.392 ft.; 9.196 ft. 9. A farmer wishes to build a granary containing 1920 cu. ft., whose dimensions are in the proportion of 5, 6, and 8; what are the dimensions? Ans. 10 ft.; 12ft.; 16 ft. 10. In digging Mr. Fisk's cellar, the length being 6 times, and the width twice the depth, 324 loads of earth were removed; what are the dimensions? Ans. 54 ft.; 18 ft.; 9 ft. 11. I have two cubical boxes, one of which will exactly |
