题目已知c/a+b+a/b+c+b/c+a=1,求c~2/a+b+a~2/b+c+b~2/c+a的值．解法 1：巧换“1”．由已知条件可知a+b+c≠0,否则a+b=-c,b+c=-a,c+a=-b,代入条件等式左边．得-3,矛盾．故可将“1”换为“a/a+b+c+b/a+b+c+c/a+b+c”,条件等式变为c/a+b+a/b+c+b/c+a=a/a+b+c+b/a+b+c+c/a+b+c．移项、整理,得 a(1/b+c-1/a+b+c)+b(1/c+a-1/a+b+c)+c(1/a+b-1/a+b+c)=O．每个括号内通分、化简．再从它们中提出1/a+b+c,得1/a+b+c(a~2/b+c+b~2/c+a+c~2/a+b)=O,从而得 The title is known as c/a+b+a/b+c+b/c+a=1, find the value of c~2/a+b+a~2/b+c+b~2/c+a. Solution 1: Skillfully change “1”. From the known conditions, we know that a+b+c≠0, otherwise a+b=-c, b+c=-a, and c+a=-b, substituting to the left of the conditional equation. -3, contradiction. Therefore, “1” can be replaced by “a/a+b+c+b/a+b+c+c/a+b+c”, and the conditional equation becomes c/a+b+a/b+c. +b/c+a=a/a+b+c+b/a+b+c+c/a+b+c. Transfer and sort, get a(1/b+c-1/a+b+c)+b(1/c+a-1/a+b+c)+c(1/a+b-1/ a+b+c)=O. Each parenthesis passes and simplifies. Then propose 1/a+b+c from them to get 1/a+b+c(a~2/b+c+b~2/c+a+c~2/a+b)=O. Get