Re: [請益] 一題高中數學
※ 引述《bbeeaauuttyy (8899)》之銘言：
: ※ 引述《ShockIdle (新的開始)》之銘言：
: 2的話 這題就找的到解了
: 以下是google到的解法~
: [2^(1/3)+1]^3 = 2+3*4^(1/3) + 3*2^(1/3) + 1
: = 3[4^(1/3) + 2^(1/3) + 1]
: = 3/[2^(1/3) -1]
: 以上這一步 最後一段我看不懂 誰能解答一下@@
: 得
: [2^(1/3)-1] = 3/[2^(1/3)+1]^3 = [4^(1/3)- 2^(1/3) +1]^3 /9
: [2^(1/3) -1]^(1/3) = [ 4^(1/3) -2^(1/3) +1] / 9^(1/3)
: = (4/9)^(1/3)- (2/9)^(1/3) + (1/9)^(1/3)
: 故 a+b+c = (4/9)-(2/9)+(1/9) = 1/3
令x=2^(1/3), x^3=2
(x+1)^3=x^3+3x^2+3x+1=3(x^2+x+1)
兩邊同乘以(x-1)
(x-1) (x+1)^3=3(x^3 -1)=3(2-1)=3
x-1=3/(x+1)^3
(x-1) ^(1/3)=[3^(1/3)]/(x+1)
=[3^(1/3)](x^2-x+1)/(x+1) (x^2-x+1)
=[3^(1/3)](x^2-x+1)/(x^3+1)
=[3^(1/3)](x^2-x+1)/3
=…
=(4/9)^(1/3)+(-2/9)^(1/3)+(1/9)^(1/3)
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