Page:Elementary algebra (1896).djvu/107

89 Rh When n is even, this expression becomes xn — yn, or zero. Therefore xn — yn is divisible by x+ y when n is even. (III.) xn + yn is never divisible by x— y. Here the substitution of y for x in the expression xn+yn gives xn + yn = yn + yn = 2yn.

As this expression is not zero, xn+ yn is never divisible by x—y.

(IV.) xn + yn is divisible by x + y when n is odd.

Here the substitution of —y for x in the expression xn+ yn gives xn - yn = (-y )n - yn When n its odd, this expression becomes — yn + yn, or zero. Therefore xn+ yn is divisible by x+ y when n is odd.

The results of the present article may be conveniently stated as follows:

(i.) For all positive integral values of n, xn — yn= (x — y) (x^{n -1}+x^{n -2}y +x^{n -3}y2 + \ldots +y^{n -1}), (ii.) When n is odd, xn +yn = (x+ y) (x^{n -1} - x^{n -2}y +x^{n -3}y2 - \ldots +y^{n -1}), (iii.) When n is even, xn — yn = (x^{n -1} - x^{n -2}y +x^{n -3}y2 - \ldots -y^{n -1}),

107. We shall now discuss some cases of greater difficulty, and also show how certain expressions of frequent occur- rence, which are not integral, may be separated into factors. The student may omit this portion of the chapter until reading the subject a second time.

Ex. 1. Resolve a^9 — 64 a3 — a^6 + 64 into six factors.

The expression

= a3 (a6 — 64) —(a3 — 64)

=(a3— 64) (a3 — 1)

= (a3 + 8) (a3 — 8)(a3 — 1)

=(a+2)(a2 —2a+4)(a — 2)(a2+ 2a+4)(a—1) (a2 + a+ 1).