答案是(ab-1)(a²-b²-1)
原式=ab(a²-b²)+a²＋b²＋ab＋1
=ab(a+b)(a-b)+(a+b)²-2ab+ab+1
=(a+b)(a²b-ab²+a+b)-ab+1
=(a+b)[a(ab-1)-b(ab-1)]-ab+1
=(a+b)[(ab-1)(a-b)]-ab+1
=(a+b)[(ab-1)(a-b)]-(ab-1)
=(ab-1)(a²-b²-1)
好累啊,自己做的,