初二希望杯因式分解(x2-1)(x4+x2+1)-(x3+1)29a2-4b2+4bc-c2x2+(a+a分之1)xy+

1)原式=(x6-1)-(x3+1)2
(x3+1)(x3-1)-(x3+1)2
(x3+1)(x3-1-x3-1)
-2(x3+2)
2）9a2-4b2+4bc-c2
9a2-(2b-c)2
(3a-2b+c)(3a+2b-c)
3)原式=x2+axy+xy/a+y2
x(x+ay)+y/a(x+ay)
(x+y/a)(x+ay)
4）25
a5+a4b+a4+a+b+1=0
a4(a+b+1)+a+b+1=0
(a4+1)(a+b+1)=0
因为3a+2b=1 所以原式等于(a4+1)(4a+3b)=0
因为a4+1大于零 所以4a+3b=0
解得a=3 b=-4