5. Seek how often the divisor may be had in the dividend, and place the result in the quotient; then multiply the divisor by this last quotient figure, placing the product under the dividend. 6. Multiply the former quotient figure, or figures by the square of the last quotient figure, and that product by 30, and place the product under the last; then under these two products place the cube of the last quotient figure, and add them together, calling their sum the subtrahend. 7. Subtract the subtrahend from the dividend, and to the remainder bring down the next period for a new dividend; with which proceed in the same manner, till the whole be finished. NOTE. If the subtrahend (found by the foregoing rule) happens to be greater than the dividend, and consequently cannot be subtracted therefrom, you must make the last quotient figure one less; with which find a new subtrahend, (by the rule foregoing) and so on until you can subtract the subtrahend from the dividend. EXAMPLES. 1. Required the cube root of 18399,744. 18399,744(26,4 Root. Ans. 8 2x2=4×300-1200) 10399 first dividend. 7200 6×6=36×2=72×30=2160 6×6×6= 216 9576 1st subtrahend. 26×26=676×500⇒205800)823744 2d dividend. 811200 4x4-16x26-416×30 12480 64 4X4X4= 823744 2d subtrahend. NOTE. The foregoing example gives a perfect root; and if, when all the periods are exhausted, there happens to be a remainder, you may annex periods of cyphers, and continue the operation as far as you think it necessary. 4,394 2,000 Answers. 59 85 346 RULE II. 1. Find by trial, a cube near to the given number, and call it the supposed cube. 2: Then, as twice the supposed cube, added to the given number, is to twice the given number added to the supposed cube, so is the root of the supposed cube, to the true root, or an approximation to it. 52,7 $,09 3. By taking the cube of the root thus found, for the supposed cube, and repeating the operation, the root will be had to a greater degree of exactness. 2,000 given number.. 2 4,000 4,32+ ,546 ,088+ 4968 EXAMPLES. Let it be required to extract the cube root of 2. Assume 1,3 as the root of the nearest cube; then1,3x1,3X1,5-2,197 supposed cube.. Then, 2,197 2 6,197 As 6,394 : : 1,3 B 1,2599 root, which is true to the last place of decimals; but might by repeating the operation, be brought to a greater exactness. 2. What is the cube root of 584,277056 ? Ans. 8,56. 16* 1 3. Required the cube root of 729001101? Ans. 900,0004 QUESTIONS, Shewing the use of the Cube Root. 1. The statute bush-1 contains 2150,425 cubic or solid inches. I demand the side of a cubic box, which shall contain that quantity? 2150,425-12.907 inch. Ans. Note-The solid contents of similar figures are in proportion to each other, as the cubes of their similar sides or diameters. 2. If a bullet 3 inches diameter, weigh 4lb. what will a bullet of the same metal weigh, whose diameter is 6 inches ? : 216: 3X3X3=27 6×6×6=216 As 27: 4lb. 32lb. Ans. 3. If a solid globe of silver, of 3 inches diameter, be worth 150 dollars; what is the value of another globe of $ silver, whose diameter is six inches ? 3X3X3=27 6×6×6=216 As 27: 150 :: 216: $1200. Ans. The side of a cube being given, to find the side of that cube wich shall be double, triple, &c, in quantity to the given cube. RULE. Cube your given side, and multiply by the given proportion between the given and required cube, and the cube root of the product will be the side sought. 4. If a cube of silver, whose side is two inches, be worth 20 dollars; I demand the side of a cube of like silve whose value shall be 8 times as much? 2×2×2=8 and 8x8-643/64-4 inches. Ans. 5. There is a cubical vessel, whose side is 4 feet; I demand the side of another cubical vessel, which shall contain 4 times as much? 4×4×4-64 and 64×4-256 3/256-6,349+ft. Ans. che cas whe cas 40 30 JAY Dex 4 whi the the the tra be re ches at the bung diameter, is ordered to make a cask of the same shape, but to hold just twice as what will be the bung diameter and length of the cask? 40×40×40×2=128000 then 32x32x32x2=65536 and 128000=50,3+ l 6553640,3+bung A General Rule for Extracting the Roots of all Po RULE. 1. Prepare the given number for extraction, by ing off from the unit's place, as the required root di 2. Find the first figure of the root by trial, and sul its power from the left hand period of the given num 3. To the remainder bring down the first figure i next period, and call it the dividend. 4. Involve the root to the next inferior power to which is given, and multiply it by the number den the given power, for a divisor. 5. Find how many times the divisor may be ha the dividend, and the quotient will be another figu the root, 6. Involve the whole root to the given power, and tract it (always) from as many periods of the given ber as you have found figures in the root. 7. Bring down the first figure of the next period t remainder for a new dividend, to which find a new sor, as before, and in like manner proceed till the w be finished. Nore. When the number to be subtracted is gre than those periods from which it is to be taken, the quotient figure must be taken less, &c. EXAMPLES. 1. Required the cube root of 135796,744 by the al general method. 135796744(51,4 the root. 75)107 dividend. 132651 2d subtrahend, 7803) 314572d dividend. 135796744 3d subtrahend. 5x5x375 first divisor. 514×514×514-135796744 third subtrahend. 3. Required the sursolid, or fifth root of 6436343. 6436343)23 root. $2 2×2×2×2×5-80)323 dividend. NOTE. The roots of most powers may be found by the For example: suppose a 12th power be given; the square root of that reduces it to a sixth power: and the square root of a sixth power to a cube. EXAMPLES. 3. What is the biquadrate, or 4th root of 19987173876? 4. Extract the square, cubed, or 6th root of 12230590 5. Extract the square, biquadrate, or 8th root of 72138 |