急
问题描述：
急
x^3-3x^2+6xy-12y^2+8y^3
x^2-2xy+y^2-2x+2y+1
a^2(a+1)-b^2(b+1)
(a-4)^2+(b-4)^2+2(ab-8)
(x^2+x+1)(x^2+x+2)-6
当x-y=1时,求x^4-xy^3-x^3y-3x^2y+3xy^2+y^4的值
有些用分组分解法(写分组过程),有些用十字相乘法.
答
x^3-3x^2+6xy-12y^2+8y^3
=(x+2y)(x^2-2xy+4y^2)-3(x^2-2xy+4y^2)
=(x^2-2xy+4y^2)(x+2y-3)
x^2-2xy+y^2-2x+2y+1
=(x-y)^2-2(x-y)+1=(x-y-1)^2
a^2(a+1)-b^2(b+1)=a^3-b^3+(a^2-b^2)
=(a-b)(a^2+ab+b^2+a+b)
(a-4)^2+(b-4)^2+2(ab-8)
=a^2+b^2+2ab-8(a+b)+16
=(a+b-4)^2
(x^2+x+1)(x^2+x+2)-6
=(x^2+x+1)^2+(x^2+x+1)-6
=(x^2+x+4)(x^2+x-1)
当x-y=1时,
x^4-xy^3-x^3y-3x^2y+3xy^2+y^4
=(x^4+y^4-2x^2y^2)-(xy^3+x^3y-2x^2y^2)-3xy(x-y)
=(x+y)^2-xy-3xy
=(x-y)^2
=1