5. Seek how often the divisor may be had in the divi. dend, 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 sub. trahend, (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. 18599,744(26,4 Root. Ans. 8 9x2=4x300=1200 10399 first dividend. 7200 6x6=36X2=72X30=2160 6x6x6= 216 9576 1st subtrahend. 26x26=676X500=20S800)825744 21 dividend. 811200 4X4=16x26=416x30= 12480 4X4X4= 64 823744 2d subtrahend. Nore.-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 operatioci as far as you think it necessary. Answers. 2. What is the cube root of 205379 ? 59 3. Of 614125? 85 4, Of 41421736? 346 5. Of 146363,183 ? 52,7 6. Of 29,503629 ? 3,097. Of 80,763 ? 4,32+ 8. Of ,162771336 ? ,546 9. Of ,000684134 ? ,088+ 10. Of 122615327232 4968 RULE II. 1. Find ly 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. 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. EXAMPLES. Let it be required to extract the cube root of %. Assume 1,3 as the root of the nearest cube; then 1,3x1,3x1,5=2,197=supposed cube. Then, 2,197 2,000 given number.. 2 2 : : As 6,394 6,197 1,3 1,2399 root, whịch 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. end 16* 3. Required the cube root of 729001101: Ans. 900,0004 che cas Whe cas 40 32 QUESTIONS, ing Shewing the use of the Cube Root. 1. The statute bugliel.contains 2150,495 cubic or solid inches. I demand the side of a cubic box, which shall contain that quantity ? 32150,125=12,907 inch. Ans. Note.---The solid contents of similar figures are in proportion to each other, as the cubes of their šunilar sides or diameters, 2. If a bullet S inches diameter, weigh Alb. what will a bullet of the same metal weigh, whose diameter is 0'. inches ? 3X3X3=27 6x6x6=216 As 27 : Alb. : 216: 321b. Ans. 3. If a solid globe of silver, of 3 inches (liameter, bę worth 150 dollars; what is the value of another globe of silver, whose diameter is six inches ? 3X3X3=27 6x6x6=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. re 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 silves, whose value shall be 8 times as much? 2x2x2=8 and 8X8=64 3/64=4 inches. Ans. 4x4x4=64 and 64x4=2563256=6,349+ft. Ins. ches at the bung diameter, is ordered to make another General Rule for Extracting the Roots of all Powers. RULE. 1. Prepare the given number for extraction, by points ing off from the unit's place, as the required root directs. 2. Find the first figure of the root by trial, and subtract its power from the left hand period of the given number. $. To the remainder bring down the first figure in the pext period, and call it the dividend. 4. İnvolve the root to the next inferior power to that which is given, and multiply it by the number denoting the given power, for a divisor. 5. Find how inany times the divisor may be had in the dividend, and the quotient will be another figure of the root, 6. Involve the whole root to the given power, and subtract it (always) from as many periods of the given number az you have found figures in the root. 7. Bring down the first figure of the next period to the remainder for a new dividend, to which find a new divisor, as before, and in like manner proceed till the whole be finished. Note. When the number to be subtracted is greater than those periods from which it is to be taken, the last quotient figure must be taken less, &c. EXAMPLES. 1. Required the cube root of 135796,744 by the above general method. i 5X5X3=75 first divisor. 514x514x514–195796744 third subtrahend 3. Required the sursolid, or fifth root of 6436343. 6436343)23 root. 2x2x2x2x5=80) 323 dividend. 23x23x23x23x23 -6436343 subtrahend. Note.The roots of most powers may be found by the square and cube roots only; therefore, when any even power is given, the easiest method will be (especially in a very high power) to extract the square root of it, which reduces it to half the given power, then the square root of that power reduces it to half the same power; and so on, till you come to a square or a cube. 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 cabe. LXAMPLES. 3. What is the biquadrate, or 4th root of 19987173376? Ans. 376. 4. Extract the square, cubed, or-6th root of 12230590 464. Ans. 48. 5. Extract the square, biquadrate, or 8th root of 72158 95789338336. Ains. 96. |