原式=(a+b)-(bc+ac)-(ab+1)+(abc+c)=(a+b)-(a+b)c-(ab+1)+(ab+1)c=(a+b)(1-c)-(ab+1)(1-c)=(1-c)(a+b-ab-1)=(1-c)[(a-ab)+(b-1)]=(1-c)[(1-b)a-(1-b)]=(1-c)(1-b)(a-1)=(a-1)(b-1)(c-1)