If, therefore, a quantity of 4 terms can be made to appear in either of these forms, it must be a perfect cube, or it consists of three equal binomial factors. Again, 27a3+27a2+9a+1 takes the form And it therefore equals (3a+1)3. (3a)3+3.(3a)2.1+3 (3a). 12+13, +(a+2b)2 (c+d) + (a+2b) (c+d) +27 (c+d)3 8 a3+3a2b2+3ab1+b6; 8x3-36x2y+54xy2 − 27y3 ; 1992 1000 abc3-30a6b+c2+ & a3b2c- 1000; (a+b)3+6 (a+b)2 c+12 (a+b) c2+8c3; }}}} a°—¥ a®b3+ #a3b® -36;125(x+y3)3+375xy (x+y); (3x2+y2) 81 84. When a quantity of four terms is equal to the sum or difference of two cubes, it may be broken into factors, as in the following examples: Ex. 1. a3-3a2+3a-7 = (a33-3a2+3a-8)+1 = = = (a-2)3+13 (a-2+1) (a-22-a-2.1+1) x+12x+y+48x2y2-61y3 = (x2+4y)*— (5y)3 Ex. 2. = (x2-y) (x+13x2y+61y2), After the necessary reductions. Examples XLIII. 6423-48x2+12x+26; 2x3+3x2+3x+1; 85. It may be found by actual multiplication, that— a3+b3c3-3abc = (a+b+c) (a2+b2+c2-ab-ac-bc). Hence we derive the method of resolving into two factors a quantity made up of the sum of the cubes of three quantities, and three times the product of the same three quantities taken negatively. a3+b3c33+3abc Ex. 1. = = Ex. 2. a3-b3c3-3abc = = Ex. 3. Ex. 4. a3+b3+ (−c)3 — 3ab (— c) (a+b−c) (a2+b2+c2¬ab+ac+bc). a3+ (− b3) + (— c)3 — 3a (—b) (−c) (a-b-c) (a2+b2+c2−ab−ac+bc). a3+b3−3ab+1; x3+y3−6xy+8; x3+y3+3xy−1; 8a3 — b3 — c3 — 6abc; 64 (x3y3+x3z3— y3z3+3x2y2z2). 27a6-10a3+1 = (3a2)3+ (2a)3+13−3 (3a2) (2a).1 x3у3+y3z3-x6z3+3x1y3z2; 8x3y3-27y3z3+64x3z3+72x2y2z2; 86. By actual multiplication, it may be seen that— a2b2+a2d2+b2c2+c2d2 = (a2+c2) (b2+d2) (1) Also, a2b2+a2d2 — b2c2 — c2d2 And a2b2-a2d2 - b2c2+c2d2 Hence it appears that when each of the four terms of an expression is the square of the product of two of four letters, each letter occurring twice only, and the signs are either all the same, or two of them positive and two negative, it may be broken into two, three, or four factors, according as it is of the form (1), (2), or (3). The following examples fall under one of the above forms ; Examples XLV. a2c2+9a2d2+4b2c2+36b2d2; x2y2+16b2x2+a2y2+16a2b2; it appears that a quantity consisting of the squares of three terms and twice the product of every two of them, is a perfect square, and can therefore be resolved into two equal factors. Note. The squares must all have the positive sign, and the products must either be all positive, or two of them must be negative and the other positive. Examples XLVI. x2+y2+4z2+2xy+4xz+4yz; 9x1 — 6x2y2-6x2+y1+2y2+1; 49a2b2+9a2c2+81b2c2-42a2bc-126ab2c+54abc2; 88. An expression of the form— a2+b2+mc2+2ab+2nac±2nbc, can sometimes be resolved into factors; for it equals and if (nm) be a square number or quantity, the expression is the difference of two squares. Similarly it may be shown that— a+b+c+2a2b2+2a2c2-2b2c2 (a2+b2+2bc+c) (a2+b2-2bc+c2). = And (b2+c2-a2)2; a+b+c+2a2b2+2a2c2+2b2c2 = (a2+b2+c2)2. Hence it appears that a quantity consisting of six terms of the same form as in (1) may always be resolved into factors, whatever be the signs of the last three terms. Resolve the following into two or four factors. Examples XLVIII. a1+4b+c-4a2b2-2a2c2-4b2c2; a++4b+c+4a2b2-2a2c2-4b2c2; a4+4b+c+4a2b2+2a2c2-4b2c2; a*+4b+c+4a2b2+2a2c2+4b2c2. Let all the following examples be taken with the same variations of sign as have been given in the above example. 2x2y2+2x2x2+2y2z2 — xa— y1—z^; 2x2y2+2x2+2y2−x1—y1 −1; + + + y1 24 x2y2 x2z2 y2z2 24a2b2x2y2-30a2c2x2x2-40b2c2y2z2 — 9a1x1 — 16b*y*—25c1z1; p2q2 p24.2 + q2p2 8a2b2c4 2a2b+c2 2a4b2c2 + CHAPTER XI. FRACTIONAL AND NEGATIVE INDICES, AND SURDS. 90. In such expressions as a3, (a+b)3, xTMy", &c., the small figure, or index, at the right hand of each quantity denotes the number of factors in it; that is |