계산:(b-c)/(a&\#178;-ab-ac+bc)-(c-a)/(b²-bc-ab+ac)+(a-b)/(c²-ab-bc+ab)

계산:(b-c)/(a&\#178;-ab-ac+bc)-(c-a)/(b²-bc-ab+ac)+(a-b)/(c²-ab-bc+ab)

∵a²-ab-ac+bc=(a-b)(a-c)b²-bc-ab+ac=(b-a)(b-c)c²-ac-bc+ab=(c-a)(c-b)∴원문=(b-c)/[(a-b)(a-c)]+(c-a)/[(b-a)(b-c)]+(a-b)/[(c-a)(c-b)]=[(a-c)-(a-b)]/[(a-b)(a-c)]+[(b-a)-(b-c)]]/[(b-c)]/[(b-c)]...