(x+y-z)^2-(x-y+z)^2=?

根据公式(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac公式展开：
得到(x^2+y^2+z^2=2xy-2yz-2xz)-(x^2+y^2+z^2-2xy-2yz+2xz)
合并同类项,最后得到4xy-4xz

原式=（x+y-z+x-y+z）（x+y-z-x+y-z）
=2x*2y
=4xy

X

(x+y-z)^2-(x-y+z)^2
=[(x+y-z)+(x-y+z)][(x+y-z)-(x-y+z)]
=(2x)(2y-2z)
=4xy-4xz

(x+y-z)^2-(x-y+z)^2=(（x+y）-z)^2-(（x-y）+z)^2=(x+y)^2-2(x+y)z+z^2-((x-y)^2+2(x-y)z+z^2)=x^2+y^2+2xy-2xz-2yz+z^2-x^2-y^2+2xy-2xz+2yz-z^2=4xy-4xz

我告诉你方法，你自己算一下，试子化为：[(x+y)-z]的平方-[(x-y)+z]的平方，
再用完全平方公式打开，打开以后(x+y)的平方和(x-y)的平方同样打开，
再消去相同的项，得结果
解题的关键是把3个项其中的其中两个看作一个

原式=[（x+y-z)+(x-y-z）]×[（x+y-z)-(x-y-z）]
=[x+y-z+x-y-z]×[x+y-z-x+y+z]
=(2x-2z)×2y
=4xy-4yz

(x+y-z)^2-(x-y+z)^2
=(x+y-z+x-y+z)(x+y-z-x+y-z)
=2x*(2y-2z)
=4xy-axz

原式=（x+y-z+x-y+z）（x+y-z-x+y-z） =2x*2y =4xy

(X+Y-Z)^2-(X-Y+Z)^2=(X+Y-Z)(X+Y-Z)-(X-Y+Z)(X-Y+Z) =(X^2+XY-XZ+XY+Y^2-YZ-XZ-YZ+Z^2 )-(X^2-XY+XZ-XY+Y^2-YZ+XZ- -YZ+Z^2 ) =(X^2+Y^2+Z^2+2XY-2XZ-2YZ)-(X^2+Y^2+Z^2-2XY+2XZ-2YZ) =4XY-4XZ

原式=（x+y-z+x-y+z）（x+y-z-x+y-z） =（2x）（2y-2z） =4xy-4xz