Factorization

y^12 - 4096
= (y^6)^2 - 64^2
= (y^6 - 64)(y^6 + 64)
= [ (y^3)^2 - 8^2 ] (y^6 + 64)
= (y^3 - 8)(y^3 + 8) (y^6 + 64)
= (y - 2)(y^2 + 2y + 4)(y^3 + 8) (y^6 + 64)
= (y - 2) (y^2 + 2y + 4) (y + 2) (y^2 - 2y + 4)(y^6 + 64)
= (y - 2) (y^2 + 2y + 4) (y + 2) (y^2 - 2y + 4) [(y^2)^3 + 4^3]
= (y - 2) (y^2 + 2y + 4) (y + 2) (y^2 - 2y + 4) (y^2 + 4)(y^4 - 4y^2 + 16)

y^12-4096
=(y^4)^3-16^3
=(y^4-16)((y^4)^2+y^4*16+16^2)
=(y^4-16)(y^8+16y^4+256).


p,s, i cannot solve(y^8-16y^4+256)!!!!!

y^12 - 4096
=y^12 - (4*1024)
=y^12 - (4*4*256)
=y^12 - (4*4*4*64)
=y^12 - (4*4*4*4*4*4)
=y^12 - 4^6
=y^12 - 2^12
=(y^6)^2 - (2^6)^2
=(y^6+2^6) (y^6-2^6)
=(y^6+64) (y^3+2^3) (y^3-2^3)
=(y^6+64) (y+3) (y^2 - 2y + 4) (y -3) (y^2 + 2y + 4)
=(y + 3) (y - 3) (y^2 - 2y + 4) (y^2 + 2y + 4) (y^6 + 64)

2007-12-22 13:40:52 補充：
The last two steps should be corrected as=(y^6 64) (y 2) (y^2 - 2y 4) (y - 2) (y^2 2y 4) =(y 2) (y - 2) (y^2 - 2y 4) (y^2 2y 4) (y^6 64)

2007-12-22 13:43:57 補充：
(y^6 64) can be further factorized into ( y^2 4)(y^4 - 4y^2 16)

y^12-4096
=y^12-2^12
=(y^6)^2-(2^6)^2
=(y^6+2^6)(y^6-2^6)