一道初二数学代数题目

a(1/b+1/c)+ a/a +b(1/c+1/a) + b/b +c(1/a+1/b) + c/c = -3 + 1 + 1 + 1
即
(a+b+c)(1/a+1/b+1/c) = 0
a+b+c = 0 或 1/a+1/b+1/c = 0
通分 (ab+bc+ac)/abc = 0 因为abc 不等于 0 所以ab+bc+ac = 0
(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ac = 1
a+b+c = 1,-1
所以 a+b+c = 0, 1, -1

因为a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=-3
所以a^2*b+a^2*c+b^2*a+b^2*c+c^2*a+c^2*b = -3abc
又(a+b+c)(a^2+b^2+c^2) = a^3+b^3+c^3+a^2*b+a^2*c+b^2*a+b^2*c+c^2*a+c^2*b = a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)
若a+b+c 不等于 0，则ab+bc+ca = 0,从而a^2+b^2+c^2+2ab+2bc+2ca = 1,(a+b+c)^2 = 1,a+b+c = 1或-1
若a+b+c = 0,代入a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=-3也满足。
所以a+b+c = -1或0或1