1.ax⑵y⑵-az⑵
问题描述：
1.ax⑵y⑵-az⑵
2.(b-c)-2a(c-b)
3.m⑵(m-1)-4(1-m)⑵
4.4a⑵-(a⑵+1)⑵
5.1-a⑵+ab-4分之1b⑵
6.x⑶-x⑵y-xy⑵+y⑶
7.（x⑵+4x)⑵-8(x⑵+4x)+16
8.(x+2)(x-2)-4y(x-y)
答
1.ax⑵y⑵-az⑵ =a(x^2y^2-z^2)=a(xy+z)(xy-z)2.(b-c)-2a(c-b) =(b-c)+2a(b-c)=(2a+1)(b-c)3.m⑵(m-1)-4(1-m)⑵ =(m-1)(m^2-4m+4)=(m-1)(m-2)^24.4a⑵-(a⑵+1)⑵ =(2a+a^2+1)(2a-a^2-1)=-(a+1)^2(a-1)^25.1-a⑵+ab-4...