3 is evidently not a common divisor of the proposed quantities, and may therefore be rejected. 30 x − 30 = 30 (x-1); reject the factor 30; Ex. 3. Required the G. C. M. of 36a6 - 18a5 - 27 a1 + 9a3 and 27a5b2-18a4b2 - 9a3 b2. Here 36a18a-27 a1+9a39a3 (4a3-2 a2 - 3a+1), and 37ab2-18 a1b2 - 9a3b2 = 9a3b2 (3a2 - 2a - 1); .. 9a3 is a factor of the G. C. M., and b2 in the latter quantity may be rejected. 2 a 3 may be easily seen not to be a common divisor of the proposed quantities, and may therefore be rejected: .. a1 is the G. c. M. of 4a3 – 2a2 3a + 1 and 3a2 – 2a — 1. Hence 9a3 (a-1) is the G. c. M. required. NOTE 5. INVOLUTION. If an algebraical expression consist of any number of parts connected together by + or, its square may readily be found by the following Rule: Square each part, and multiply twice that part into the sum of the several parts that come after; the sum of all the results so obtained will be the square of the whole quantity. Thus, Ex. 4. (1 + x + 2x2 + 3 x3 + 4x2 + 5x3 + 2 ... = 1 + 2(x+2x2 + 3x3 + 4x1 +5∞3 +6x+...) + x2+ 2x (2x2 + 3∞3 + 4x2 + 5x5 +.....) +4x2+4x2 (3x3 + 4x1...) +9x+6x3 (4x1 +.....) + = 1 + 2x + 5x2+10x3 + 18 x2 + 30 x3 + 47x6 +... It is useful to commit to memory the following formulæ, viz., the cube of any binomial quantity; (A + B)3 = A3 + B3 + 3 AB (A + B) . ..... (A – B)3 = A3 – B3 − 3 AB (A – B) ... Ex. 1. Required the cube of a+b+c. (a+b+c)3 = (a+b+c)3 ...... (1.) ....... (a + b)3 + c3 + 3 (a + b) c (a + b + c), by (1) a3 + b3 + 3ab (a + b) + c3 + 3 (ac + bc) (a + b + c). 3 Ex. 2. Required the cube of Va+a-Va-x. The following forms also ought to be committed to memory: Ex. 1. A3 ÷ B3 = (A + B) . (42 – AB + B2) ... (3) A3 - B3 (AB) . (A2 + AB + B2) ... (4). = x2 + ya = (x3 + y1) . (x − √ xy + y). Ex. 2. x − y} = (x3 − y3). (x + √ xy + y). ył THE square root of any number may be found by using the common Rule for extracting the square root until one more than half the number of digits in the root is obtained; then the rest of the digits in the root are determined approximately by Division. Let N represent the number whose square root is required; the part of the root found by the common Rule; a that is, N-a2 divided by 2a will give the rest of the square digits, x2 has 2n at most. But, by the supposition, a has 2n+1 digits at least; therefore Hence it appears, that if n + 1 digits of a square root are obtained by the common Rule, n digits more may be correctly obtained by Division only. Ex. Required the square root of 2 to 6 places of decimals. 2.0000... (1·414 1 24) 100 96 281) 400 281 2824) 11900 2828) 604000 (213 5656 3840 2828 10120 8484 1636 .. the root required is 1414213...... When a cube root consists of 2n + 2 digits, and n +2 have been found by the ordinary rule, the rest can be found by dividing by the trial divisor. Let a + b be the cube root, where a consists of n + 2 digits, and n cyphers, a3 + 3a2b+3ab2 + b3 the quantity whose root is required. Then after a has been found, we have remainder = 3a2b+3ab2 + b3; .. quotient b+ − + = b3 |