把(a+b-c)(a-b+c)+(b-a-c)²因式分解

(a+b-c)(a-b+c)+(b-a-c)²
=[a+(b-c)][a-(b-c)]+[b-(a+c)]²
=a²-(b-c)²+b²-2b(a+c)+(a+c)²
=a²-b²+2bc+b²-c²-2ab-2bc+a²+2ac+c²
=2a²-2ab+2ac
明白请采纳,
有新问题请求助,-(b-c)²是怎么算的？-(b-c)²的结果是什么？-(b-c)²=-(b²-2bc+c²)=-b²+2bc-c²