多項式因式分解

1
a^2 - 2ab + b^2 - 6(a-b)+9
= (a – b)^2 – 6(a-b) + 9
= (a – b –3)^2

2
2p^6 - 128
= 2 (p^6 – 64)
= 2 (p^3 – 8) (p^3 + 8)
= 2 (p – 2) (p^2 + 2p + 4) (p^3 + 8)

3
y^6 - 7y^3 – 8
= (y^3 – 8) (y^3 + 1)
= (y – 2) (y^2 + 2y + 4) (y^3 + 1)

(1)
a^2 - 2ab + b^2 - 6(a-b)+9
=(a-b)^2 - 6(a-b) +9
=((a-b)-3)^2
=(a-b-3)^2

(2)
2p^6 -128
=2(p^6 - 64)
=2(p^2*3 - 4^3)
=2(p^2 - 4)((p^2))^2 - 4(p^2) - 4^2)
=2(p^2 - 2^2)((p^2)^2 - 2(2)(p^2) - ((2)^2)^2)
=2(p - 2)(p + 2)(p^2- 2^2)^2
=2(p - 2)(p + 2)((p - 2 )(p + 2))^2
=2((p - 2)(p + 2))^3

(3)
y^6 - 7y^3 - 8
=(y^3)^2 - 7(y^3) - 8
=((y^3) - 8)((y^3)+1)
=((y^3 -(2)^3)(y^3 + (1)^3)
=(y - 2)(y^2 + 2y +4)(y + 1)(y^2 - y + 1)

2007-09-23 19:06:11 補充：
Corr:(3) y^6 - 7y^3 - 8=(y^3)^2 - 7(y^3) - 8=((y^3) - 8)((y^3) 1)=((y^3 -(2)^3)(y^3 (1)^3)=(y - 2)(y^2 2y 4)(y 1)(y^2 - y 1)=(y - 2)(y 2)^2(y 1)(y^2 - y 1)=(y^2 - 4)(y 2)(y 1)(y^2 - y 1)=(y^2 - 4)(y^2 3y 2)(y^2 -y 1)

1) a^2-2ab+b^2-6(a-b)+9
=(a-b)^2-6(a-b)+9
=(a-b-3)^2


3) y^6-7^3-8
=(y^3-2^3)(y^3+1^3)
=(y-2)(y^2+2y+4)(y+1)(y^2-y+1)
=(y^2-y-2)(y^2+2y+4)(y^2-y+1)



not sure

2007-09-23 18:42:52 補充：
2) 2p^6 - 128 =2(p^6-64) =2(p^3-8)(p^3+8) =2(p-2)(p^2+2p+4)(p+2)(p^2-2p+4) =2(p^2-4)((p^4+8p^2+16) =2(p^2-4)(p^2+4)^2