参考答案

(1)由分配律与向量积的反交换律
(a+b+c)×c+(a+b+c)×b-(b-c)×a
=(a+b+c)×(b+c)-(b-c)×a
=ax(b+c)+a×(b-c)=a×2b (因为(b+c)×(b+c)=0)
=2a×b．
(2)(a-2c)·[(a-b)×(a-b-c)]
=(a-2c)·[(a-b)×(a-b)-(a-b)×c]
=(a-2c)·(-a×c+b×c)
-a·(-a×c)+a·(b×c)+2c·(a×c)-2c·(b×c)
=a·(b×c)．
最后一步的理由是由向量积的定义a⊥(-a×c)，c⊥(a×c)，c⊥(b×c)，因此a·(-a×c)=0．c·(a×c)=0．c·(b×c)=0．
(3)