【正确答案】[证明]  (bcd)a-(cda)b+(dab)c-(abc)d=(bcd)a+(dab)c-[(cda)b+(abc)d](bda)c-(bdc)a-[(acd)b-(acb)d]=(b×d)×(c×a) -(a×c)×(b×d)=-(b×d)×(a×c)+(b×d)×(a×c)=0.