已知（x-y-3）^2+|xy-4|=0,求x^2+y^2 （x+y）^2 x^3y-2x^2y^2+xy^3的值

（x-y-3）^2+|xy-4|=0
∴﹛x-y-3=0
xy-4=0
即﹛x-y=3
xy=4
x^2+y^2
=x²+y²-2xy+2xy
=(x-y)²+2xy
=3²+2×4
=17
（x+y）^2
=x²+2xy+y²
=17+2×4
=25
x^3y-2x^2y^2+xy^3
=xy(x²-2xy+y²)
=xy(x-y)²
=4×3²
=36