Interesting Maths Q14 (Normal)

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/701112310055813873417710.jpg


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

This is the answer for question 1.


圖片參考：http://imgcld.yimg.com/8/n/HA00901242/o/701112310055813873417710.jpg

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

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