polynomial`

1st method
Treat x6=(x), y6=(y) and apply the formula (a-b)=(a+b)(a-b)
2nd method
Treat x6=(x), y6=(y) and apply the formula a-b=(a+b)(a-ab+b)

By formula (a^2 - b^2) = (a-b) (a b)
(a^3 - b^3) = (a-b) (a^2 ab b^2)
(a^3 b^3) = (a b) (a^2 - ab b^2)
First way (by using formula)
x^6 - y^6
= (x^3)^2 - (y^3)^2
= (x^3 - y^3) (x^3 y^3)
= (x - y) (x^2 xy y^2) (x y)(x^2 - xy y^2)
Second way (By completing square or cube method)
x^6 - y^6
= x^6 - 2x^3y^3 y^6 2x^3y^3 - 2y^6
= (x^3 - y^3)^2 2 y^3(x^3 - y^3)
= (x^3 - y^3) [(x^3 - y^3) 2y^3]
= (x^3 - y^3) (x^3 y^3)
= [(x^3 - 3x^2y 3xy^2 - y^3) 3x^2y - 3xy^2]
[(x^3 3x^2y 3xy^2 y^3) - 3x^2y -3xy^2]
= [(x-y)^3 3xy(x-y)] [(x y)^3 - 3xy(x y)]
= (x-y) [(x -y)^2 3xy] (x y) [(x y)^2 - 3xy]
= (x-y) (x^2 xy y^2) (x y) (x^2 - xy y^2)


2008-07-01 16:01:40 補充：
( re-coding problem - so the '+' symbols are missing)
Revised the second method

x^6 - y^6

= x^6 - 2x^3y^3 + y^6 + 2x^3y^3 - 2y^6

= (x^3 - y^3)^2 + 2 y^3(x^3 - y^3)

= (x^3 - y^3) [(x^3 - y^3) + 2y^3]

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

continued

2008-07-01 16:01:49 補充：
= [(x^3 - 3x^2y 3xy^2 - y^3) 3x^2y - 3xy^2] [(x^3 3x^2y 3xy^2 y^3) - 3x^2y -3xy^2]

= [(x-y)^3 3xy(x-y)] [(x+y)^3 - 3xy(x + y)]

= (x-y) [(x -y)^2 3xy] (x+ y) [(x+y)^2 - 3xy]

= (x-y) (x^2+ xy+ y^2) (x+ y) (x^2 - xy +y^2)

First method
x^6-y^6
=(x^2)^3-(y^2)^3
=(x^2-y^2)(x^4+x^2y^2+y^4)
Second Method
(x^6)-y^6
=(x^3)^2-(y^3)^2
=(x^3+y^3)(x^3-y^3)
=(x+y)(x^2-xy+y^2)(x-y)(x^2+xy+y^2)
=(x+y)(x-y)(x^4+x^3y+x^2y^2-x^3y-x^2y^2-xy^3+x^2y^2+xy^3+y^4)
=(x+y)(x-y)(x^4+x^2y^2+y^4)

2008-07-01 15:52:30 補充：
Sorry, first method, we can factorise as
(x+y)(x-y)(x^4+x^2y^2+y^4)

x6-y6
=(x3)2-(y3)2
=(x3-y3)(x3+y3)
=(x-y)(x2+xy+y2)(x+y)(x2-xy+y2)