# 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.

To Rvy:

Is there any mistake in your Q2 last line?

***Q1 last line.

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

In Q1 last line:

55(±√5)=±275√5

??!!

Note that x>0.

Actually there's a simpler way.

### 4 個解答

• 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.

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

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• This is the answer for question 1.

資料來源： mainly self
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• 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

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• 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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