will be +, and the odd powers -: as is evident from multiplication.* * Any power of the product of two or more quantities is equal to the same power of each of the factors multiplied together. And any power of a fraction is equal to the same power of the numerator divided by the like power of the denominator. Also, am raised to the nth power is amn; and - am raised to the nth power is amn, according as n is an even or an odd number. EXAMPLES FOR PRACTICE. Ans. 8a. 1. Required the cube, or third power, of 2a2. 3. Required the cube, or third power, of Ans. 16ax4. 2 x2y. 3 5. Required the fourth power of a + x; and the fifth power Ans. a2 + 4a3x + 6a2x2 + 4ax3 +x*, and a of a - y. 5a2y + 10a3y2 - 10a2y + 5ay - y5. RULE 2.-A binomial or residual quantity may also be readily raised to any power whatever, as follows : 1. Find the terms without the coefficients, by observing that the index of the first, or leading quantity, begins with that of the given power, and decreases continually by 1, in every term to the last; and that in the following quantity, the indices of the terms are 1, 2, 3, 4, &c. 2. To find the coefficients, observe that those of the first and last terms are always 1; and that the coefficient of the second term is the index of the power of the first: and for the rest, if the coefficient of any term be multiplied by the index of the leading quantity in it, and the product be divided by the number of terms to that place, it will give the coefficient of the term next following. Note. The whole number of terms will be one more than the index of the given power; and when both terms of the root are +, all the terms of the power will be +; but if the second term be -, all the odd terms will be +, and the even terms; or, which is the same thing, the terms will be + and - alternately.* * The rule here given, which is the same in the cases of integral powers as in the binomial theorem of Newton, may be expressed in general terms, as follows: m-1 m-2 a m-1 m-2, b2+m b2-m EXAMPLES. 1. Let a + x be involved, or raised to the fifth power. Here the terms, without the coefficients, are, a, ax, x2, a2x3, ax, x5. And the coefficients, according to the rule, will be 5×4 10 × 3 10 ×25×1 1, 5, 2 or 1, 5, 10, Whence the entire fifth power of a +x is 1, a + 5a^x+10ax2 + 10a2x3 + 5ax2 + x3. 3. Let a - x be involved, or raised, to the sixth power. Here the terms, without their coefficients, are, a, ax, ax, ax3, α2x, ax, x. And the coefficients, found as before, are, 1, 6, 6×5 15 X 420 × 3 15 ×26×1 , , 4 20, 15, , 5 6; or 1, 6, 15, Whence the entire sixth power of a - x is a- 6x + 15a2x2 - 20α2x2 + 15a2x2 - 6ax5 + x. 3. Required the fourth power of a + x, and the fifth power of a - x. Ans. a2 + 4x + 6a2x2 + 4ax3 + x2, and a 5a*x + 10a3x2 - 10a2x2 + 5ax - x. 5. Required the sixth power of a + b, and the seventh power Ans. a + 6a2b+ 15a2b2 + 20ab3 + 15a2b+ of a y. 6ab5 + bo, and a7 - 7ay +21a2y2 - 35ay + 35ay - 21a2y + 7ay - y". 6. Required the fifth power of 2 + x, and the cube of a-bx+c. Ans. 32 + 80x80x2+40x2 + 10x + x5, and a3+3a2c + 3ac2 + c2 - 3a2bx - 6acbx — 3c2bx + 3ab2x2 + 3cb2x2 - bx3. EVOLUTION. EVOLUTION, or the extraction of roots, is the reverse of involution, or the raising powers; being the method of finding the square root, cube root, &c., of any given quantity. CASE I. To find any root of a simple quantity. RULE. Extract the root of the coefficient for the numeral &c. which formulæ will also equally hold when m is a fraction, as will be more fully explained hereafter. It may also be farther observed, that the sum of the coefficients in every power, is equal to the number 2 raised to that power. Thus 1+1 = 2, for the first power; 1+2+1=4=22, for the square; 1+3+3+1 =8=23, for the cube, or third power; and so on. part, and the root of the quantity subjoined to it for the literal part; then these joined together, will be the root required. And if the quantity proposed be a fraction, its root will be found by taking the root both of its numerator and denominator. Note. The square root, the fourth root, or any other even root, of an affirmative quantity, may be either + or -. Thus √ a2 = + a or -a, and b2 = + bor - b, &c. But the cube root, or any other odd root, of a quantity, will have the same sign as the quantity itself. Thus, =a; - a2 = - a; and a = a, &c.* It may here, also, be farther remarked, that any even root of of a negative quantity is unassignable. Thus, ✓ a cannot be determined, as there is no quantity, either positive or negative, (+ or -), that, when multiplied by itself, will produce - a2. EXAMPLES. 1. Find the square root of 9x2; and the cube root of 8x3. Here 9x = √9 × √ x2 = 3 x x = 3x. And / 8x3 = 8 x 3 x = 2 × x = 2x. 2. It is required to find the square root of Ans. Ans. and the 4c3 4. It is required to find the cube root of - 125x. 3. It is required to find the square root of 4ax. Ans. 2ax. Ans. 5ax2. 5. It is required to find the 4th root of 256ax. Ans. 4ax3. 4a+ 6. It is required to find the square root of 2a 7. It is required to find the cube root of * The reason why + a and -a are each the square root of ar is obvious, since, by the rule of multiplication, (+a)×(+a) and (-a) X (-a) are both equal to a2. And for the cube root, fifth root, &c., of a negative quantity, it is plain, from the same rule, that (-a)×(-a)×(-a)=-a3; and (-as) x (+2)=-a5. And consequently a3=-a, and 5 a5-a. To extract the square root of a compound quantity. RULE 1.-Range the terms, of which the quantity is composed, according to the dimensions of some letter in them, beginning with the highest, and set the root of the first term in the quotient. 2. Subtract the square of the root thus found, from the first term, and bring down the two next terms to the remainder for a dividend. 3. Divide the dividend, thus found, by double that part of the root already determined, and set the result both in the quotient and divisor. 4. Multiply the divisor, so increased, by the term of the root last placed in the quotient, and subtract the product from the dividend; and so on, as in common arithmetic. EXAMPLES. 1. Extract the square root of x* - 4x3 + 6x2 - 4x + 1. x2 - 4x + 6x2 - 4x + 1 (x2 - 2x + 1 Ans. x2 - 2x + 1, the root required. 2. Extract the square root of 4a2 + 12ax+13a2x2 + ax36 + x2. 4a2 + 12a3x + 13a2x2 + 6ax + x2 (2a2 + Зах + x2 4a 4a3+3ax) 120x + 13a2x2 12a3x + 9α2x 4a3+6ax + x2) 4α2x2 + 6ax3 + x2 * |