the root, which is 18ab, together with the square of the last term 36, which is 962. Multiplying then the complete divisor by 36, and subtracting the product from the last remainder, nothing is left. Hence the required cube root is 2a+3b. This result may be easily verified by multiplication. 206. If the root contains three terms, as a+b+c, we may put a+b=m. Then (a+b+c)3=(m+c)3=m3+3m2c+3mc2+c3. If we proceed as in the last example, we shall find a+b, and we subtract its cube from the given polynomial. There will then remain 3m2c+3mc2+c3, which may be written (3m2+3mc+c2)c. We perceive that 3m2 will be the new trial divisor to obtain We then complete the divisor by adding to it 3mc+c2. Let it be required to find the cube root of 8a6—36ba3+ €6b2a*—63b3a3 +33b1a2—9b3a+bo. 8a-36ba3+66b2a*—63b3a3+33b*a2—9b3a+bo(2a2—3ba+b2 The first term of the root is 2a2, and subtracting its cube, the first term of the remainder is -36ba5, which, divided by 3 times the square of 2a2, gives -3ba for the second term of the root. Complete the divisor as in the last example, and multiply it by -3ba. Subtracting the product from the last remainder, the first term of the second remainder is 12b2a1. To form the new trial divisor, we take three times the square of the part of the root already found, viz., 2a2—3ba. Divide the first term of the remainder by 12a1, and we obtain b2 for the last term of the root. We now complete the divisor by adding to it three times the product of the third term by the sum of the first two terms, and also the square of the last term. Multiplying the divisor thus completed by l2, we find the prod uct equal to the last remainder. Hence the required cube root is 2a2-3ba+b2. 207. Hence, for extracting the cube root of a polynomial, we derive the following RULE. 1st. Arrange the terms according to the powers of some one letter; take the cube root of the first term, and subtract the cube from the given polynomial. 2d. Divide the first term of the remainder by three times the square of the root already found; the quotient will be the second term of the root. 3d. Complete the divisor by adding to it three times the product of the two terms of the root and the square of the second term. 4th. Multiply the divisor thus increased by the last term of the root, and subtract the product from the last remainder. 5th. Take three times the square of the part of the root already found for a new trial divisor, and proceed by division to find another term of the root. 6th. Complete the divisor by adding to it three times the product of the last term by the sum of the first two terms, and also the square of the last term, with which proceed as before till the entire root has been obtained. We may dispense with forming the complete divisor according to the rule if each time that we find a new term of the root we raise the entire root already found to the third power, and subtract the cube from the given polynomial. EXAMPLES. 1. What is the cube root of ao-6a5+15a*—20a3+15a2 6a+1? Ans. a2—2a+1. 2. What is the cube root of 6x5-40x3+x+96x-64? Ans. 3. What is the cube root of 18x*+36x2+24x+8+32x3+ 2+6x5? Ans. Ans. 4. What is the cube root of 365+66-563-1+36? G 5. What is the cube root of 8x6-36x5+66x1-63x3+33x29x+1? 6. What is the cube root of 8x +48αx5+60a2x2-80a3x390a*x2+108a5x-27α®? 7. What is the cube root of 8x6—36ax3+102a2x2-171a3x3 +204a2x2-144a3x+64α®? Cube Root of Numbers. 208. The preceding rule is applicable to the extraction of the cube root of numbers; but a difficulty in applying it arises from the fact that the terms of the powers are all blended together in the given number. They may, however, be separated by attending to the following principles: 1st. For every three figures of the cube there will be one figure in the root, and also one for any additional figure or figures. Thus, the cube of 1 is 1 the cube of 1 is 1 1,000 1,000,000 1000" 1,000,000,000 Hence we see that the cube root of a number consisting of from one to three figures will contain one figure; the cube root of a number consisting of from four to six figures will contain two figures; of a number from seven to nine figures will contain three figures, and so on. Hence, if we divide the number into periods of three figures, commencing at units' place, the number of periods will indicate the number of figures in the cube root. 209. 2d. The first figure of the root will be the cube root of the greatest cube number contained in the first period on the left. For the cube of tens can give no significant figure in the first right-hand period; the cube of hundreds can give no figure in the first two periods on the right; and the cube of the highest figure in the root can give no figures except in the first period on the left. Let it be required to extract the cube root of 438976. This number contains two periods, indicating that there will 702x3=14700 70×6×3= 1260 438,976 (70+6, the root. 343,000 95976 62= 36 95976 complete divisor, 15996 be two places in the root. Let a be the value of the figure in the tens' place, and b of that in the units' place. Then a must be the greatest multiple of 10 which has its cube less than 438000; that is, a must be 70. Subtract the cube of 70 from the given number, and the remainder is 95976. This remainder corresponds to 3a2b+3ab3 +63, which may be written (3a2+3ab+b2)b. Divide this remainder by 3a2, that is, by 14700, and the quotient is 6, which is the value of b. Complete the divisor by adding to it 3ab, or 1260, and b2, or 36. The complete divisor is thus found to be 15996, which, multiplied by 6, gives 95976. Subtracting, the remainder is zero, and we conclude that 70+6, or 76, is the required cube root. For the sake of brevity the ciphers may be omitted, provided we retain the proper local values of the figures. If the root consists of more than two places of figures, the method will be substantially the same. Let it be required to extract the cube root of 279,726,264. 279,726,264 (654 216 108 90 63726 25 11725 58625 12675 780 5101 264 Having found 65, the cube root of the greatest cube contained in the first two periods, we bring down the last period, and have 5101264 for a new dividend. We then take three times the square of the root already found, or 12675, for a partial divisor, whence we obtain 4 for the last figure of the root. We then complete the divisor by adding to it three times the product of 4 by 65, and the square of 4, regard being paid to the proper local values of the figures. The complete divisor is thus found to be 1275316, which, multiplied by 4, gives 5101264. Hence 654 is the required cube root. 16 1275316 5101 264 210. Hence, for the extraction of the cube root of numbers, we derive the following .RULE. 1st. Separate the given number into periods of three figures each, beginning at the units' place. 2d. Find the greatest cube contained in the left-hand period; its cube root is the first figure of the required root. Subtract the cube from the first period, and to the remainder bring down the second period for a dividend. 3d. Take three hundred times the square of the root already found for a trial divisor; find how many times it is contained in the dividend, and write the quotient for the second figure of the root. 4th. Complete the divisor by adding to it thirty times the product of the two figures of the root, and the square of the second figure. 5th. Multiply the divisor thus increased by the last figure of the root; subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. 6th. Take three hundred times the square of the whole root now found for a new trial divisor, and by division obtain another figure of the root. 7th. Complete the divisor by adding to it thirty times the product of the last figure by the former figures, and also the square of the last figure, with which proceed as before, continuing the operation until all the periods are brought down. It will be observed that three times the square of the tens, when their local value is regarded, is the same as three hundred times the square of this digit, not regarding its local value. In applying the preceding rule, it may happen that the product of the complete divisor by the last figure of the root is greater than the dividend. This indicates that the last figure of the root was taken too large, and this happens because the divisor is at first incomplete, that is, too small. In such a case we must diminish the last figure of the root by unity, until we obtain a product which is not greater than the dividend. EXAMPLES. 1. Find the cube root of 163667323. Ans. 547. |