（-1/3a^2b^2)^2(-2ab^2)^3÷(1/9a^4b^4)^3 [(x+y)^2-(x+y)(x+y)]÷2y
问题描述：
（-1/3a^2b^2)^2(-2ab^2)^3÷(1/9a^4b^4)^3 [(x+y)^2-(x+y)(x+y)]÷2y
（-1/3a^2b^2)^2(-2ab^2)^3÷(1/9a^4b^4)^3
[(x+y)^2-(x+y)(x+y)]÷2y
因式分解：
x^3y-6x^2y^2+9xy^3
(a^2+4b^2)2-16a^2b^2
答
（-1/3a^2b^2)^2(-2ab^2)^3÷(1/9a^4b^4)^3
=1/9a^4b^4×(-8a³b^6)÷(1/729的12次方b的12次方)
=-648/a的5次方b²
[(x+y)^2-(x+y)(x+y)]÷2y
=0÷2y
=0
x^3y-6x^2y^2+9xy^3
=xy(x²-6xy+9y²)
=xy(x-3y)²
(a^2+4b^2)2-16a^2b^2
=(a²+4b²+4ab)(a²+4b²-4ab)
=(a+2b)²(a-2b)²