m m 6m m m 22. a3x3-3a2bx2ñ y2m — b3y&m + 3 ab2xnym +3a2 cx2nzn m 2m 4m 3n - 6 ab cxn y2mzn + 3b2 cy1mzn + c3z3n + 3 ac2x z2n 23. Verify the correctness of the following expressions by extracting the cube roots: How can we derive one of these formulas from the other? 24. With the assistance of one of the series in 23, calculate the value of 137 to seven decimal places. 3 3 Here 31/37 = 130 V/1 10/11, etc. Put x=1 in 23. 10009 1 0 25. Similarly find the value of 128 to six decimal places. THE CUBE ROOT OF ARITHMETICAL NUMBERS 296. The first step in finding the cube root of numbers expressed by figures is to point the number off into periods. Therefore, the cube root of any number between 1 and 1000, i. e., of any number which has one, two, or three figures, is a number of one figure; the cube root of any number between 999 and 1000000, that is, of any number which has four, five, or six figures, is a number of two figures, and so on. Hence, if a point is placed over every third figure in any number, beginning with the units, the number of points will show the number of figures in the cube root. 297. If the cube root of a number contains decimal places, the number itself will have three times as many. Thus, if .5 is the cube root of a number, the number will be (.5) (.5) (.5) = .125; and if 2.3 is the cube root of some number, the number will be 12.167. Hence, if the given cube number has decimal places, it will have three times as many decimals as its cube root. Therefore, if the given number has decimal figures, and a point is placed over the units' figure, and over every third figure to the right and left of it, then the number of points in the decimal part of the number will indicate the number of decimal places in its cube root. If the given number is not a perfect cube, ciphers may be annexed, and a value of the root may be found as near to the true value as one chooses by repeating the process for finding the cube root. 298. Some examples in the extraction of the cube root of arithmetical numbers are now given, the rule being derived from the rule for finding the cube root of a polynomial. EXAMPLE I. Find the cube root of 2628072. The pointing shows that the root will consist of three figures. The largest cube root in the first period, 2, is 1; by subtracting The figure 1 is the cube of 1 the remainder 1628072 is obtained. in the hundreds' place, so the trial divisor is 3(100)2=30000, which corresponds to 3a, example I, 285. The trial divisor 30000 may apparently be contained three, four, or five times in 162807, but it will be found on trial that 4 and 5 are too large; therefore 3 will be the second figure of the root. The complete trial divisor 3a2+3ab+b will in this example be 3(100)+3(100×30)+(30)2= 39900, which, multiplied by 30 gives 1197000, and subtracted from 1628072 leaves the second remainder 431072. The new trial divisor (295) 3(x+y) is in this case 3 (100+ 30)2 = 3 (130)o = 50700. 431072 divided by 50700 gives 8 in the third figure of the root. The complete divisor 3(x + y)2 + 3(x + y)z + z2 (295) will, in this case, be 3(100+30)2+3(100+30)8+8=53884, which multiplied by 8 and subtracted from the last remainder leaves zero. 300. By repeating the steps for finding the cube root of a perfect cube, the cube root of a number which is not a perfect cube can be found to any desired degree of approximation. Thus, find the cube root of 6.21, correct to the third decimal place. 301. If the cube root of a number consists of 2n+2 figures, in case the first n + 2 of these have been found by the usual method, the remaining n can be found by division. Let N be the number; a the part of the cube root already found, that is, the first n + 2 figures found by rule, with n zeros annexed; x the remaining part of the root. Now N-a3 is the remainder after n + 2 figures of the root represented by a a have been found; and 3a is the corresponding trial divisor. Equation (1) shows that N-a3 divided by 3a2 gives x, the remaining part of the cube root required, increased by + It can now be shown that x2 a 3 a so that, by neglecting the remainder arising from the division, x, the rest of the root required, is obtained. Thus, by hypothesis, x contains n figures, and a, n + 2 figures, 34. v 74300, 7430, 1/743, 174.3, 7.43, 10.743. 35. Having found four figures of the cube roots of the numbers in examples 3, 4, 6, 11, and 13, find three more figures in each of these roots by 2301. 36. 282429536481. 37.208827064576. 302. The nth Root of a Polynomial.-A rule for finding the nth root of a polynomial can be obtained by observing the formation of the nth power of a polynomial, n being any integral number whatever (1265). (a + b)" = a2 + n an-1b + Thus Therefore "Van +na"-1b+ = a + b. |