来自董华明的问题
【(2m-n)ˆ(2m+n)ˆ2(nˆ2+4mˆ2)ˆ2(x+y)ˆ2(x-y)ˆ2-(x-y)(x+y)(xˆ2+yˆ2)】
(2m-n)ˆ(2m+n)ˆ2(nˆ2+4mˆ2)ˆ2
(x+y)ˆ2(x-y)ˆ2-(x-y)(x+y)(xˆ2+yˆ2)


【(2m-n)ˆ(2m+n)ˆ2(nˆ2+4mˆ2)ˆ2(x+y)ˆ2(x-y)ˆ2-(x-y)(x+y)(xˆ2+yˆ2)】
(2m-n)ˆ(2m+n)ˆ2(nˆ2+4mˆ2)ˆ2
(x+y)ˆ2(x-y)ˆ2-(x-y)(x+y)(xˆ2+yˆ2)
第一道全是幂的形式么、没看懂
第二道=(x+y)(x-y)(x+y)(x-y)-(x^2-y^2)(x^2+y^2)=(x^2-y^2)(x^2-y^2)-(x^2-y^2)(x^2+y^2)=(x^2-y^2)(x^2-y^2-x^2-y^2)=-2y^2(x^2-y^2)
(2m-n)ˆ2(2m+n)ˆ2(nˆ2+4mˆ2)ˆ2没符号就是乘
第一道=(2m-n)(2m+n)(2m-n)(2m+n)(n^2+4m^2)^2=(4m^2-n^2)(4m^2-n^2)(n^2+4m^2)(n^2+4m^2)=(4m^2-n^2)^2