计算 〔x-2)*3+〔2-y)*3-〔x-y)*3

楼主的题目应该是3次方吧

〔x-2)^3+〔2-y)^3-〔x-y)*^3
=(x-2)^3+[(2-y)-(x-y)]*[(2-y)^2+(2-y)*(x-y)+(x-y)^2]
=(x-2)^3+(-x+2)*[(2-y)^2+(2-y)*(x-y)+(x-y)^2]
=(x-2){(x-2)^2-[(4-4y+y^2)+(2x-2y-xy+y^2)+(x^2-2xy+y^2)]
=(x-2)(x^2-4x+4-4+4y-y^2-2x+2y+xy-y^2-x^2+2xy-y^2)
=(x-2)(-6x+6y+3xy-3y^2)
=3(x-2)[-2(x-y)+y(x-y)]
=3(x-2)(y-2)(x-y)

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原式等于
x`3-2x`2+4x-8+(8-4y+2y`2-y`3) -(x`3-x`2*y+x*y`2-y`3)

等于 -2x`2+4x-4y+2y`2+x`2*y-x*y`2

化不下去了。。。。

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原式=(x-2)^3-(x-y)^3+(2-y)^3
=(x-2-x+y)(x^2+4-4x+x^2+y^2-2xy+x^2-xy-2x+2y)+(2-y)^3
=(y-2)(3x^2-3xy-6x+y^2+2y+4)-(y-2)^3
=(y-2)(3x^2-3xy-6x+y^2+2y+4-y^2-4+4y)
=(y-2)(3x^2-3xy-6x+6y)
=3(y-2)(x(x-y)-2(x-y))
=3(y-2)(x-2)(x-y)

〔x-2)*3+〔2-y)*3-〔x-y)*3
=〔x-2)*3-〔x-y)*3+〔2-y)*3
=(x-2-x+y)(x*2-4x+4+x*2-xy-2x+2y+x*2-2xy+y*2)+〔2-y)*3
=(y-2)(3x*2-3xy+y*2-6x+2y+4)-(y-2)*3
=(y-2)(3x*2-3xy+y*2-6x+2y+4-y*2+4y-4)
=(y-2)(3x*2-3xy-6x+6y)