數學a.maths 中四野

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

b) x^5 + 1/x^5
= (x + 1/x)^5 - 5x^4(1/x) - 10x^3(1/x)^2 - 10x^2(1/x)^3 - 5x(1/x)^4
= (x + 1/x)^5 - 5x^3 - 10x - 10/x - 5/x^3
= (x + 1/x)^5 - 5(x^3 + 1/x^3) - 10 (x + 1/x) .......from (a) x^3+1/x^3 =2
= 1^5 - 5(-2) - 10(1)
= 1 + 10 - 10
= 1

(p.s. (a+b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5 )

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

(x + x^-1)^3

= x^3 + 3X^2(x^-1) + 3x(x^-2) + x^-3

= x^3 + x^-3 + 3(x + x^-1)

therefore,
x^3 + x^-3

= (x + x^-1)^3 - 3(x + x^-1)

= 1 -3(1)

= -2


(b)
Simiarly,
(x + x^-1)^5 = 1^5 = 1

(x + x^-1)^5

= x^5 + 5x^4(x^-1) + 10x^3(x^-2) + 10x^2(x^-3) + 5x(x^-4) + x^-5

= x^5 + x^-5 + 5(x^3 + x^-3) + 10(x + x^-1)

x^5 + x^-5

= (x + x^-1)^5 - 5(x^3 + x^-3) - 10(x + x^-1)

= 1 - 5(-2) -10(1)

= 1

(a) x^3 + 1/x^3 = (x+1/x)(x^2-1+1/x^2) = 1[(x+1/x)^2 -2-1] = 1[(1)^2-3] = -2
(b) x^5 + 1/x^5 = (x+1/x)(x^4-x^2+1-1/x^2+1/x^4) = 1[(x+1/x)^4-5x^2-5/x^2-5]
=1^4 - 5(x+1/x)^2 + 5 = 1-5+5 = 1