就三道简单的因式分解题,(1)x^3-(x^2)y-xy^2+y^3(2)x^2-8xy+15y^2+2x-6y(3)已知x^2+4x-1=0,求2x^4+8x^3-4x^2-8x+1的值.（注：如x^3就是x的三次方）

1 x2(x-y)-y2(x-y)
=(x-y)(x2-y2)
=(x-y)(x+y)(x-y)
=(x-y)2(x+y)
2 (x-5y)(x-3y)+2(x-3y)
=(x-3y)(x-5y+2)
3 2x4+8x3-2x2-2x2-8x+2-1
=2x2(x2+4x-1)-2(x2+4x-1)-1
因为x^2+4x-1=0
所以 0-0-1=-1

x^3-(x^2)y-xy^2+y^3=x^3+y^2-xy(x+y)=(x+y)(x^2-xy+y^2)-xy(x+y)=(x+y)(x^2-xy+y^2-xy)=(x+y)(x-y)^2x^2-8xy+15y^2+2x-6y=(x-5y)(x-3y)+2(x-3y)=(x-3y)(x-5y+2)x^2+4x-1=0x^2+4x=12x^4+8x^3-4x^2-8x+1=2x^2(x^2+4x)...