Elements of Algebra

CHAP. III. OF ALGEBRA. 83 It remains now to multiply the first product I. II. by this second product III. IV. $a^3+2a^2b+2ab^2+b^3$ $a^3-2a^2b+2ab^2-b^3$ $a^6+2a^5b+2a^4b^2+a^3b^3$ $-2a^5b-4a^4b^2-4a^3b^3-2a^2b^4$ $2a^4b^2+4a^3b^3+4a^2b^4+2ab^5$ $-a^3b^3-2a^2b^4-2ab^5-b^6$ $a^6-b^6$ which is the product required. 279. Now let us resume the same example, but change the order of it, first multiplying the factors I. and III. and then II. and IV. together. $a+b$ $a-b$ $a^2+ab$ $-ab-b^2$ $a^2-b^2$ $a^2+ab+b^2$ $a^2-ab+b^2$ $a^4+a^3b+a^2b^2$ $-a^3b-a^2b^2-ab^3$ $a^2b^2+ab^3+b^4$ $a^4+a^2b^2+b^4$ Then multiplying the two products I. III. and II. IV. $a^4+a^2b^2+b^4$ $a^2-b^2$ $a^6+a^4b^2+a^2b^4$ $-a^4b^2-a^2b^4-b^6$ $a^6-b^6$ which is the product required. 280. We may perform this calculation in a manner still more concise, by first multiplying the Ist. factor by the IVth, and then the IId. by the IIId. $a^2-ab+b^2$ $a+b$ $a^3-a^2b+ab^2$ $a^2b-ab^2+b^3$ $a^3+b^3$ $a^2+ab+b^2$ $a-b$ $a^3+a^2b+ab^2$ $-a^2b-ab^2-b^3$ $a^3-b^3$ G 2