[a^3(c-b)+b^3(a-c)+c^3(b-a)]/[a^2(c-b)+b^2(a-c)+c^2(b-a)]=a+

真不好意思,看到题目时已经太晚了
先看a^3(c-b)+b^3(a-c)+c^3(b-a)
=a³(c-b)+b³[(a-b)+(b-c)]+c³(b-a)
=a³(c-b)-b³(b-a)-b³(c-b)+c³(b-a)
=(a³-b³)(c-b)+(c³-b³)(b-a)
=(a-b)(a²+ab+b²)(c-b)-(c-b)(c²+bc+b²)(a-b)
=(a-b)(c-b)(a²+ab+b²-c²-bc-b²)
=(a-b)(c-b)[(a²-c²)+(ab-bc)]
=(a-b)(c-b)[(a+c)(a-c)+b(a-c)]
=(a-b)(c-b)(a-c)(a+b+c)
再看a^2(c-b)+b^2(a-c)+c^2(b-a)
=a²(c-b)+b²[(a-b)+(b-c)]+c²(b-a)
=a²(c-b)-b²(b-a)-b²(c-b)+c²(b-a)
=(a²-b²)(c-b)+(c²-b²)(b-a)
=(a-b)(a+b)(c-b)-(c+b)(c-b)(a-b)
=(a-b)(c-b)[(a+b)-(c+b)]
=(a-b)(c-b)(a-c)
所以[a^3(c-b)+b^3(a-c)+c^3(b-a)]/[a^2(c-b)+b^2(a-c)+c^2(b-a)]
=[(a-b)(c-b)(a-c)(a+b+c)]/ [(a-b)(c-b)(a-c)]
=a+b+c
等式恒成立
【希望不要晚了!】
顺便问一下
是你提问的么?
我也回答了.