用简便方法计算：(x+2)^2(x^2-4x+4)^2+(x-2)^2(x^2+4x+4)^2

(x+2)^2(x^2-4x+4)^2+(x-2)^2(x^2+4x+4)^2
=[(x+2)(x^2-4x+4)]^2+[(x-2)(x^2+4x+4)]^2
=(x^3+8)^2+(x^3-8)^2
=x^6+16x^3+64+x^6-16x^3+64
=2x^6+128

(x+y)^2-(x^3+y^3)/(x+y)
=(x+y)^2-(x+y)(x^2-xy+y^2)/(x+y)
=x^2+2xy+y^2-x^2+xy-y^2
=3xy

[(x+y)^2-4xy][(x-y)^2+3xy]
=(x^2+2xy+y^2-4xy)(x^2-2xy+y^2+3xy)
=(x-y)^2(x^2+xy+y^2)
=(x-y)[(x-y)(x^2+xy+y^2)]
=(x-y)(x^3-y^3)
=x^4-x^3y-xy^3+y^4

若（a+b）^2=15,(a-b)^2=11,求a^2+b^2-2ab
错了吧
是不是求a^2+b^2-ab？
（a+b）^2=15,(a-b)^2=11
相加
a^2+2ab+b^2+a^2-2ab+b^2=15+11
2(a^2+b^2)=26
a^2+b^2=13
（a+b）^2=15,(a-b)^2=11
相减
(a+b)^2-(a-b)^2
=(a+b+a-b)(a+b-a+b)
=2a*2b=4ab=15-11
ab=1
所以a^2+b^2-ab=13-1=12

(x+2)^2(x-2)^2+(x-2)^2(x+2)^2=2*[(x+2)(x-2)]^2=2(x^2-4)^2
=2x^4-16x^2+32

(x+y)^2-(x^2-xy+y^2)(x+y)/(x+y)=x^2+2xy+y^2-(x^2-xy+y^2)=3xy

(x-y)^2[(x+y)^2-xy]=[(x-y)(x+y)]^2-(x-y)^2*xy=(x^2-y^2)^