In extracting the square root of a number, where there is a decimal, the pointing off into periods must proceed both ways from the decimal point; and if, after all the figures of the given number are brought down, there should be a remainder, the process of extraction may be carried on to any degree of accuracy, by continually annexing two ciphers to the remainder. 59. The rule for the extraction of the cube root of a compound quantity, may be obtained in a similar manner, by observing how the terms of the root are to be derived from those of the power. The cube of a+b is a3+3a2b+ 3ab2+b3, which is composed of the cubes of the two terms, together with three times the square of the first into the second, and three times the square of the second into the first. If, therefore, the terms be arranged according to the powers of a, the cube root a of the first term a3 will be the first term of the root. If the cube of a be subtracted from the given power, there will remain 3a2b+3ab2+b3; the first term of which, divided by three times the square of a, the part of the root already found, will give b the second term of the root. Now, if to three times the square of the first term, three times the product of the two terms, together with the square of the last term be added, the divisor will be complete. This divisor being multiplied by the last term, and the product subtracted will leave no remainder in this particular example. If there were a remainder, then the other terms of the power would be annexed to it, and the process would proceed as before, until all the terms were exhausted. Ex. 1. Required the cube root of a3+3a2b+3ab2+63. a3+3a2b+3ab2+b3(a+b a3 3a2+3ab+b2)3a2b+3ab2+b3 3a2b+3ab2+b3 Ex. 2. Required the cube root of x-6x+15x1 —20x3+ 15x2-6x+1. x6-6x5+15x1-20x3+15x2-6x+1(x2-2x+1 ენ 3x-6x+4x2-6x+15x1—20x3 3x-12x+12x2+3x2-6x+13x-12x+15x2-6x+1 1) Or, 3x2-12x3+15x2-6x+1)3x21—12x3+15x2-6x+1 60. By a similar process, we are enabled to extract the cube root of any number. Ex. 1. Let it be required to find the cube root of 16974593. By omitting the superfluous ciphers, making a small change upon the factors of the two first of those quantities which form the divisors, and bringing down three figures at a time, the above process will then be changed into the following more convenient form : As every example is susceptible of a like condensation, the following general rule may be given. Divide the given number into periods of three figures each, beginning from the place of units. Find the cube root (2) of the greatest cube number (8) contained in the first period (16), which will be the first figure of the root. Subtract the cube of Το this figure from the first period, and to the remainder (8) annex the next, (974). Divide this number by 300 times the square of that figure (2) of the root already found, and the quotient* (5) will be the next figure of the root. 300 times the square of the first figure (2), add 30 times the product of the two figures (2x 5) of the root now found, together with the square of the latter (5); multiply this sum (1525) by the last figure (5) of the root, and from the dividend (8974) subtract the product (7625). To the remainder (1349) annex the next period (593), and proceed as before. There are still some considerations, by which the labour prescribed in the preceding rule, may be abridged. Thus, to multiply by 300, is the same as to multiply by 3, and then annex two ciphers; and, by 30, to multiply by 3, and annex one cipher. Again, referring to the example at the beginning of the present number, it is manifest that the a employed in finding the second divisor, is equal to the a+b employed in *It must here be remarked, that this quotient is in general much less than the true quotient, because of the additions which the divisor is afterwards to receive. Thus, 1200 is contained 7 times in 8974, but if 7 were taken in place of 5, it would be found that the resulting quantity, when multiplied by 7, would exceed 8974. finding the first; the 3a2, therefore, of the second divisor, will be equal to 3(a+b)2 of the first = 3a2+6ab+3b2, which is greater than the first complete divisor 3a2+3ab+b2 by Sab+262; if, therefore, 3ab+262 be added to the first divisor, the sum will be the first part of the succeeding one. This, then, in regard to the numbers, will be found by adding the three last lines together, observing to reckon the middle one twice. Thus, -152500+2× 2500+30000= 187500, the first part of the next divisor; and, in the same way, may the first part of every succeeding divisor be found. The reason of dividing the given number into periods, of three figures each, is because the cube of a number can never have more than three times as many places as its root, and at the least but two less; consequently, there will always be just as many places in the root, as there are pe riods of three figures each in the given number, the left hand period consisting in some instances of one or two figures. By attending to the preceding remarks, the process of the extraction of the cube root of any number may be exhibited in the following condensed form; where, for the sake of conciseness, the sign + is used instead of the word annex. Ex. 2. Required the cube root of 51097851963712. The fourth root of a number, or algebraic quantity, may be found by first extracting the square root, and then the square Thus, 256√256 = √16=4. root of that square root. In the same manner, the sixth root may be found by extracting the cube root, and then the square root of that cube root. Thus, 61. The roots of algebraic quantities may be expressed by fractional indices. = a. Square root of a2 = a = a; the cube root of a6a2 a2 m In the same manner, the nth root of aman, from which it appears, that the roots of quantities are found by dividing the index of the quantity by the index of the root. Hence, square root of a or a = a; cube root of a or Jaa; cube root of a2 or a2-a; the fifth root of a or a3a. In like manner, the square root of a+ or √a+=(a+x); the cube root of (a+x)2 or 3⁄4√(a+x)2 = (a+x)3, &c. 3 62. Any quantity raised to the power, whose index is 0, is equal to unity. m an and will be the same as which is 1, because a quanti an am am ty divided by itself unity; hence ao = 1. 63. The powers and roots of quantities may be expressed by negative indices. |