5. To the remainder annex the next period, and use the last perfect divisor as an imperfect divisor for the new dividend, and proceed as before directed. JNote 1.—When the right hand figure is omitted, to multiply by 3, put a cipher in its place. If the right hand place be a cipher, it is not to be omitted. 1. What is the cube root of 5122 - Explanation.—When only three places of figures are given, the root is found by trial. he same method must be pursued with the first period of a larger number. Ans. 8. produce 216? Ans. 6. Manner of finding the perfect divisor. 6 0 three times the whole root. 2 3 whole root. 1 8 O 1 2 O Product of 3 1 3 8 O times the whole - + root except 3 9 square of the right hand fig. into the whole yoot 1 3 8 9 perfect divisor. o : * JNote 2. —The first imperfect divisor is hundreds when considered in connexion with the dividend, the two right hadd figures of the dividend should therefore be omitted in ascertaining how many times it can be taken from the dividend It will also be proper to observe here, that any imperfect divisor is so much smaller than the perfect one, that the next figure to be obtained in the root, will not be so large as might at first be supposed by comparing it with the dividend. Ans, 32. 8. Extract the cube root of 795.07. N Root, 43. 9. Required the cube root of 175616. Ans. 56. 10. What number equals the cube root of 3007634 o Ans. 67. 11. What number multiplied twice into itself, will make 343000 7 Ans. 70. 12. What number raised to the third power equals 512000 7 Ans. 80. 13. Of what number is 729000 the cube 2 Ans. 90. JNote 3.—When a remainder occurs, the operation may be continued by annexing periods of three ciphers each. The figures obtained in the root after annexing the ciphers, will be decimals. When the cube root of a vulgar fraction is required, the cube root of the numerator will be the numerator of the root, and the cube root of the denominator, the denominator, if the operations can be performed without a remainder to either ; if not, the vulgar fraction must be changed to a decimal. Illustration of the principles on which the foregoing method of extracting the cube root, is founded. The 3d power of a number is a number expressing the cubical or solid inches, feet, &c. contained in any body in the form of a cube ; the cube root of that number is the length of one side. (See page 245.) When the solid contents of a cubical body are given, or any number of cubical inches, feet, &c. to be arranged in the form of a cube, to find the length of one side we extract the cube root. To illustrate that operation by the rule, will give a result corresponding with what has been said of the cube root, let us suppose we have 2744 cubical blocks of wood, each containing 1 solid inch. If these blocks be placed together, so that the number of blocks in the length, breadth, and thickness shall be equal, the pile would be in the form of a cube, and the number of blocks it is high, wide, or long, that is, the number of inches, will be the cube root. In 2744 there are two periods, consequently there must be two places of figures in the root. Operation. 2 7 4 4 ( 1 4 The greatest cube in 2, I the left hand period, is 1, -*- and the root 1; or the great3 (1 7 4 est cube in 2000 is 1000, and 1 7 4 the cubc root of 1000 is 10,0e4 36 —4-- noted by the first figure in the root. By subtracting 1000. from the whole number, we suppose blocks enough taken to make a cubical pile, whose length, breadth, or thickness is 10 inches, since 10 multiplied into itself twice, produces 1000, the number of blocks already taken from the whole number, and which form the supposed cube, whose side is 10 inches. In fig. 1, let the square A represent one side, or Fig. 1. face, of the cubical pile made by the 1000 blocks; then the side a b, which is * equal to each of the other sides, will be 10 inches in length. 1744 blocks still remain to be added to the pile A. To keep the pile in the form of a cube, the additions of the remainb ing blocks must be made to three sides. The next object is to find how many blocks will be required to cover those three sides, one block, or inch thick. This we shall do by ascertaining how many square inches there are on the three sides. 1st part of operation. (1 0 The square of a b, or 10 2 7 4 4 1 0 0 0 inches, is 100 square inches, the number of square 3 o 0 ) 1 7 4 4 inches on the whole face - A ; and 3 times 100 are 300, the number of square inches on the three faces to which the three additions are to be made. 300 is used as an imperfect divisor, since as many times as 300 can be taken out of the remaining number of blocks, so many blocks or inches in thickness, can be added to 3 faces, each equal to A. It appears by the first part Fig. II. . of the operation, that 300 can be taken 5 times from 1744, but by adding to one side, the other sides are increased, so that it will require more than 300 blocks to cover 3 faces each larger than A. In fig. 2, let the part A represent the face of the cube formed with the 1000 blocks, the same as in fig. 1. Now if we add several thicknesses to the side a c, the sides d b and on b d are each larger than the part A. A perfect divisor must be a number expressing the number of square inches on the three sides to which the additions are to be made, after the sides are increased, 300 is therefore too small for a perfect divisor. Let us suppose the real divisor can be taken 4 times from the remaining number of blocks, then the whole length of one side of the block, after the additions are made, must be 14 inches. In fig. 2, the part as represents one layer of blocks added to the side c a and extending to s. If the additions can be made 4 inches thick, scnust be 4 inches long, since c a is 10; and since it is 10 inches from a to the bottom of the pile, the layers a must be 14 inches long, and 10 inches wide, |