Page:Elementary algebra (1896).djvu/108

90 Rh Ex. 2. a (a- 1) x^3 -(a-b-1) xy - b (b + 1) y2 = {ax-(b+1)y}{(a - 1)x + by}.

108. From Ex. 2, Art. 59, we see that the quotient of a^3 + b^3 + c^3 — 3 abc by a + b + c is a^2 + b2 + c2 — bc — ca — ab.

Thus a^3 + b^3 + c^3 — 3 abc = (a + b + c)(a^2 + b2 + c2 — bc — ca — ab).

This result is important and should be carefully remembered. We may note that the expression on the left consists of the sum of the cubes of three quantities a, b, c, diminished by three times the product abc. Whenever an expression admits of a similar arrangement, the above formula will enable us to resolve it into factors.

Ex. 1. Resolve into factors a^3 — b^3 + c^ + 3 abc.

a^3 + b3 + c3 + 3 abc = a3 + (— b)3 + c3 - 3 a (- b) c = (a — b + c) (a2 + b^2 + c^2 + bc - ca + ab), — b taking the place of b in formula (1).

Ex. 2. x^3-8y^3-27 -18xy = x3 + (-2y)3+ (- 3)^3 - 3x (- 2y)(- 3) = (x - 2y -3)(x^2 + 4y^2 + 9 - 6 y + 3x + 2xy).

109. Expressions which can be put into the form x3 \pm {1} {y3} may be separated into factors by the rules for resolving the sum or the difference of two cubes. [Art. 102.]

Ex.1. {8}{a3}—27b6

Ex 2. Resolve a2x3 - 8a2 y3 - x3 + {8} {y3} into four factors.