Interesting Maths Q14 (Normal)

1. If x-x⁻¹=1, find x¹º-x⁻¹º.

2. If a+b+c=x, ab+ac+bc=y, abc=z, find a³+b³+c³.

更新:

To Rvy:

Is there any mistake in your Q2 last line?

更新 2:

***Q1 last line.

更新 3:

If I say in Q1, x>0, do you have any simpler method?

更新 4:

In Q1 last line:

55(±√5)=±275√5

??!!

更新 5:

Note that x>0.

Actually there's a simpler way.

4 個解答

評分
  • Rvy
    Lv 6
    8 年 前
    最佳解答

    1 is more difficult than 2

    2 is just simply cubic equation

    ( a cubic equation m^3-xm^2+ym-z=0 with roots a,b and c )

    2.

    (a+b+c)^3 << in order to get the terms a^3, b^3 and c^3

    =a^3+3a^2(b+c)+3a(b+c)^2+(b+c)^3

    =a^3+3a(b+c)[a+(b+c)]+(b^3+3ab^2+3a^2b+b^3)

    =a^3+3(ab+ac)(x)+(b^3+3b^2c+3bc^2+c^3)

    =a^3+b^3+c^3+3(ab+ac)(x)+3b^2c+3bc^2

    =a^3+b^3+c^3+ 3(ab+ac+bc)(x)-3bcx +3bc(b+c)

    =a^3+b^3+c^3+3(y)(x)+3bc(b+c-x)

    =a^3+b^3+c^3+3(y)(x)+3bc(-a)

    =a^3+b^3+c^3+3xy-3z

    therefore, a^3+b^3+c^3=x^3-3xy+3z

    1.

    圖片參考:http://imgcld.yimg.com/8/n/HA00285646/o/7011123100...

    Sorry, i have no idea to calculate it faster

    2012-01-01 19:25:16 補充:

    full size:

    http://i40.tinypic.com/9tfx9w.png

  • 8 年 前

    This is the answer for question 1.

    圖片參考:http://imgcld.yimg.com/8/n/HA00901242/o/7011123100...

    資料來源: mainly self
  • 8 年 前

    But if you have this Q in exam,

    you have to prove it is Fibonacci Sequence...

    TROUBLESOME!

    ***Great discovery!

    2012-01-01 22:55:22 補充:

    Um.. gd...

    2012-01-01 23:03:25 補充:

    To CRebecca:

    Do you interested in my other "Interesting Maths Qs" ??

    Many of them are still empty! XD

  • 8 年 前

    Q1: Fibonacci Sequence

    Q2: a,b,c are roots of t³-xt²+yt-z=0

    2012-01-01 22:29:36 補充:

    let x^n= a(n)+b(n)x, then

    x^(n+1)=a(n+1)+b(n+1)x=a(n)x+b(n) (x+1)=b(n)+[a(n)+b(n)]x

    so, x^n= a(n)+a(n+1)x and {a(n)} is the Fibonacci Sequence

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