已知:x+y=3,xy=﹣8,求:
(1)x2+y2
(2)(x2﹣1)(y2﹣1).
解:(1)∵x+y=3,xy=﹣8,
∴原式=(x+y)2﹣2xy=9+16=25;
(2)∵x+y=3,xy=﹣8,
∴原式=x2y2﹣(x2+y2)+1=64﹣25+1=40