(A maths)Binomial Theorem questions~

(a)
x + 1/x = 1
(x + 1/x)^3 = 1^3
x^3 + 3x + 3/x + 1/x^3 = 1 . . . . . . . (*)
x^3 + 3(x + 1/x) + 1/x^3 = 1
x^3 + 3(1) + 1/x^3 = 1
x^3 + 3 + 1/x^3 = 1
x^3 + 1/x^3 = -2

(b)
x + 1/x = 1
(x + 1/x)^5 = 1^5
x^5 + 5x^3 + 10x + 10/x + 5/x^3 + 1/x^5 = 1 . . . . . . . (*)
x^5 + 5(x^3 + 1/x^3) + 10(x + 1/x) + 1/x^5 = 1
x^5 + 5(-2) + 10(1) + 1/x^5 = 1
x^5 - 10 + 10 + 1/x^5 = 1
x^5 + 1/x^5 = 1


(*) 這一步驟用了 Binomial Theorem 去展開

2006-11-10 16:17:42 補充：
還有一點，希望你都可以了解一下。(x + 1/x)^2 = x^2 + 2 + 1/x^2 (x + 1/x)^3 = x^3 + 3x + 3/x + 1/x^3(x + 1/x)^4 = x^4 + 4x^2 + 6 + 4/x^2 + 1/x^4... 餘此類推其實它們都有 pattern 出現，所以你再次見到 x+1/x 或 x-1/x 之類的題目(樣子相似的都唔好放過)，可以不妨將它用 Binomial Theorem 去 expand。

a) x^3+1/x^3
= (x^6 + 1)/x^3 ----------------------------- (1)

x + 1/x = 1
(x^2 + 1)/x = 1
(x^2 + 1) = x --------------------------------- (2)
(x^2 + 1)^3 = x^3
x^6 + 3x^4 + 3x^2 + 1 = x^3
x^6 + 1 + 3x^2(x^2 + 1) = x^3 --------- (3)

use (2) to sub into (3)
x^6 + 1 + 3x^2(x) = x^3
x^6 + 1 + 3x^3 = x^3
x^6 + 1 = x^3 - 3x^3 = -2x^3
(x^6 + 1)/x^3 = -2