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)
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)