RULE.-For extracting the Cube Root. 1. Separate the given number into periods of three figures each, by putting a point over the unit figure, and every third figure beyond the place of units. 2. Find the greatest cube in the left hand period, and put its root in the quotient. 1 3. Subtract the cube, thus found, from the said period, and to the remainder bring down the next period, and call this the dividend. ! 4. Multiply the square of the quotient by 300, calling it the divisor. 5. Şeek how many times the divisor may be had in the dividend, and place the result in the root; then multiply the divisor by this quos tient figure, and write the product under the dividend. 6. Multiply the square of this quotient figure by the former figure or figures of the root, and this product by 30, and place the product under the last; under all, write the cube of this quotient figure, and call their amount the subtrahend. 7. Subtract the subtrahend from the dividend, and to the remainder bring down the next period for a uew dividend, with which proceed as before; and so on, till the whole is finished. EXAMPLES 1. A man has 13824 feet of timber, in separate blocks, each containing one cubick foot, and having a desire to place : them in a cubick pile, he wishes to know the length of each side of such a pile. Ans. 24 feet. | We first ascertain what number of figures we must have in the root, by placing a period over the unit figure, and then one over the third beyond it; these periods show that the root must consist of as many figures as we have periods; they, likewise divide the dividend into its proper parts, which musi be known before we can proceed in the operation of the work. We place a period over the unit figure, and one over every third counting to the left, because no figure multiplied by its square, or used three times as a factor, will exceed three figures. | DEM.--After placing the periods, as our rule directs, we seek the greatest cube in the left hand period, 13, without regard to its local value, which we find, by the table of powers, to be 8, placing its root, 2, in the quotient for the first figure of the root, and subtracting the 8 from the left hand period. And since we have two periods placed over our given number, our root must consist of two figures; therefore the 2, the first figure of the root, must be 2 tens, then 20 must be the local value of the 2. Now, at all stages of the work, the root, reckoned according to its local value, expresses one side of a cube or square solid, formed from what has been subtracted from the given number. Then let. Figure 1. represent the cube or square solid which has been formed from what has been taken from our dividend. One side of this cube being 20 feet, as our root shows, its solid contents must be 20x20x20=8000 .solid feet, which exactly agrees with the number taken from our first period, because our first period expresses thou And it is sands, and 8 standing under thousands has a local value of so many thousands, consequently it has taken from our first period 8000, leaving 5 thousand, the remainder of the first period, to the right of which we bring down the next period, 824, leaving for our entire remainder 5824 solid feet, which forms a new Operation. dividend. And now the re maining 5824 solid feet, we 138242 must so dispose of, in enlarging the cube, as to preserve its 8 cubick forin; and in order 5824 to do this, it is evident, that the addition to the cube, A, D, must be made on three Fig. Í. sides, a, b, and c. C 20 D plain, if we dividë 5824 solid feet, by the area or surface of the three sides, a, b, and c; the quotient will be the thickB 20 ness of the addition made to each of the sides, a; b; and c. Then the next thing to be 20 done, is to form a divisor, which shall express the area E or surface of the three sides, A 2,-3, and C. And to proceed as our rule directs, we multi ply the square of the quotient 20 by 300 ; thus, 2X2=4, the 20 square of the quotient, with out regard to its local value; 400 then 4X300=1200 square feet, 20 for the divisor, which is half the area of the cube A, D, 8000 or the area of the three sides, a, b, and c. But the root, %, tnat is 2 tens, already found, is the length of one of the sides; there fore we might proceed, thus 20X20=400 square feet, the area of one side, which multiplied by 3, the number of sides, gives 100X3=1200 square feet, the area of the three sides as before ; but this last method is not so convenient in practice, therefore, we place the ciphers of both factors at the right of our multiplier 3, and call it 300 as in the first operation. We now find, that our divisor, 1200, is contained in the dividend, 5824, 4 times, consequently the last figure of our root, 4 feet, expresses the thickness of the addition made to each of the sides, a, b, and c; then multiplying 1200 by 4, must give 4800 solid feet, the number contained in the three additions, because it is the product arising from multiplying the length, breadth, and thickness of the three additions together. But on inspecting Fig. II, we find that this addition made of the three sides, falls short of completing the cube ; because there are still deficiencies in the three corners, n, n, n. Now it is evident, that the length of the solid required to fill each of these deficiencies, must be the sanie as the length of the cube before any addition was made; therefore 2, the first figure of the root, must represent the length of each of these deficiencies, because it represents the length of each side of the cube before any addition was made; but this 2 expresses 20, for it stands in the place of tens , then 20 must be the length of each of these defiOperation continued. ciencies, and ihe width and thickness of each must be 13824(24 Ropt. equal to the last quotient fi. gure, 4; therefore the solid feet 8 o necessary to fill the deficienDivisor, 1200)5824 Dividena. cies, n, n, in, may be found by multiplying 16, the square ot 4800 the last quotient figure, 4, by 960 the length of all the deficien cies. And it has been shown 64 that the length of each defiSubtrahend, 5824 ciency is 20 feet, then the length of the three deficiencies 0000 must be 60 feet; therefore 16, the product of the breadth and Fig. II. thickness of each deficiency, 20 multiplied by 60, the lengih of the whole, must give the solid contents required to fill the three deficiencies, n, n, n; thus, 16X60=960 solid feet, the solidity required to 20 fill the three deficiencies, n, n, n. Or the soliļity fequired to 20 fill the three deficiencies, may be obtained thus, 4X4=16, the product of the breadth and thickness of each deficiency, or the square of the last quotient figure, then 16X2=32, 20 which multiplied by 30, thus, 32X30=960° solid feet, the same result as before. But perhaps it would have rendered the work more plain, if we had multiplied 16, the square of the last quotient figure, or the product of the breadth and thickness of each deficiency, by 20, the length of each deficiency, which would have given the solidity required to fill one deficiency, and then have multiplied that solidity by 3, the number of deficiencies in the corners, n, , n ; thus, 16X20=326, the solidity of one deficiency, then, 320X3=960 before. And now having supplied the deficiencies, n, n, n, we find on inspecting Fig. III, that there still remains a deficiency in the corner where the three last pieces meet. Now it is plain, that this last deficiency must be a small cube, because the solid required to fill this deficiency, must be equal in length, breadth and thickness, to the last quotient figure, 4. Then by Fig. III. čubing this last quotient figure, we will have the solid feet re20 + 4 quired to fill the vacancy, thus, 4 20 4X4X4=64 solid feet, the cortents of the corner piece, which will be seen in its place in Fig. IV. 20 The number of solid feet in these several additions must be 20 added to compose the subtrahend, thus, 4800+960 +64= 5824, the number of solid feet in the subtrahend, that is, in all the additions, which, sub20 + 4 tracted from the dividend, ' leaves no remainder, and the Fig. IV. work is now finished. 24 feet. Then Fig. IV. represents the cubick pile, each side of which is 24 feet, and contains 13824 solid feet, which may be proved by cubing one of the sides of the cube, that is, the root, thus, 24 X 24 X 24=13824 solid feet, the given number, that is, multiplying the length, breadth, and thickness together, to find the solidity; or it may be proved by adding together the solid feet contained in the several 24 feet. parts, thus: Feet. 8000=contents of the first cube, Fig. I. 4800=first addition to the sides a, b, c, Fio. II. 64=3d addition to fill the corner e, e, e, Fig. IV. 13824=contents of the whole pile, Fig. IV., and 24 seet, the cube root or length of each side. 2. What is the cube root of 34645976? 34645976(326 Answer. 27 5400 Carried over. 24 5400 Brought over 22X3X30= 360 23: 8 5768 first subtrahend. 32 x300=307200)1877976 second dividend. 1843200 69-X32X30= 34560 63 216 1877976 second subtrahend. DEM.-It has been stated that the root, at all stages of the work, represents the side of a cube, formed from what has been taken from the given number. Now 3, the first figure of the root, having a local value, represents the side of a cube, each side of which must be 300, for if we raise 300 to the third power, the product will exactly agree with the number subtracted from the given number. When we have obtained two figures, 32, in the root, they likewise having a local value, represent the side of a cube formed from what has been subtracted from the dividend or given number; and each side of the cube, at this stage of the work, must be 320. And when we have obtained three figures in the root, as 326, our root, 326, must represent the side of a cube, the solid contents of which are equal to the given number. Then it is evident, that after every subtraction our root represents the side of a new cube, so that for every figure in the root, after the first, there must be additions made similar to that which was made in obtaining the second figure of the root under example ist; the reason of which is evident from the demonstration there given. 3. What is the cube root of 729? Ans. 9. 4. What is the cube root of 12167 ? Ans. 23. 5. What is the cube root of 4826809 ? Ans. 169. 6. What is the cube root of 9129329 ? Ans. 209. 7. What is the cube root of 15625000 ? Ans. 250. 8. What is the cube root of 329939371,? Ans. 691. 9. What is the cube root of 997002999 ? Ans. 999. 10. What is the cube root of 729000000 ? Ans. 900. Note.—To extract the cube root of a Vulgar Fraction, first extract the cube root of the numerator for a new numerator, and then extract the cube root of the denominator for a new denominator. Or reduce the fraction to a decimal, and extract the cube root of the decimal. 11. What is the cube root of ? Ans. . 12. What is the cube root of 115? Ans. . 13. What is the cube root of 115 ? When there are decimals in the given number, point off the wbclé numbers the same as if there were no decimals belonging to the given Ans. š. |