方程x-a-b/c+x-b-c/b+x-c-a/b=3,且1/a+1/b+1/c不等于0,求x的值

3移项过来,是-3
拆成-1-1-1
则(x-a-b)/c-1+(x-b-c)/a-1+(x-c-a)/b-1=0
(x-a-b-c)/c+(x-a-b-c)/a+(x-a-b-c)/b=0
(x-a-b-c)(1/a+1/b+1/c)=0
1/a+1/b+1/c不等于0
x-a-b-c=0
x=a+b+c