(b-c)(c-a)分之a-b +（c-a）(a-b)分之b-c +(a-b)(b-c)分之c-a
=(a-b)/(b-c)(c-a)-(b-c)/(a-c)(a-b)-(a-c)/(a-b)(b-c)
=[(a-b)²-(b-c)²-(a-c)²]/[(a-b)(b-c)(a-c)]
=(a²-2ab+b²-b²+2bc-c²-a²+2ac-c²)/[(a-b)(b-c)(a-c)]
=(-2ab+2bc+2ac-2c²)/[(a-b)(b-c)(a-c)]
=[-2b(a-c)+2c(a-c)]/[(a-b)(b-c)(a-c)]
=-2(a-c)(b-c)/[(a-b)(b-c)(a-c)]
=-2/(a-b)
∴原式的值不可能为零