∵| x + y + 1|≥0,| xy - 3 |≥0
| x + y + 1| +| xy - 3 | = 0,
∴x+y+1=0,即x+y=-1
xy=3
xy3 + x3y
=xy(x²+y²)
=yx[(x+y)²-2xy]
=3×(1-6)
=-15