题目应该是4a²b²-(a²+b²-c²)²吧=(2ab)²-(a²+b²-c²)²=(2ab+a²+b²-c²)(2ab-a²-b²+c²)=[(a+b)²-c²][c²-(a-b)²]=(a+b+c)(a+b-c)(c+a-b)(c-a+b)
4a²b²-(a²+b²-c²)²=(2ab+a²+b²-c²)(2ab-a²-b²+c²)=【(a+b)²-c²】【c²-(a-b)²】=(a+b+c)(a+b-c)(c-a+b)(c+a-b)