a、b、c为非零实数，a*a+b*b+c*c=1，a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)=-3，求a+b+c

a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=-3
a(1/b+1/c)+1+b(1/c+1/a)+1+c(1/a+1/b)+1=-3+3
a(1/a+1/b+1/c)+b(1/a+1/b+1/c)+c(1/a+1/b+1/c)=0
(a+b+c)*(1/a+1/b+1/c)=0
所以
a+b+c=0或1/a+1/b+1/c=0 但因为abc均不为0
所以a+b+c=0
a/abc+b/abc+c/abc=0/abc
(bc+ac+ab)/(abc)=0

又因为abc不等于0
所以ab+ac+bc=0

a^2+b^2+c^2=1
a^2+b^2+c^2+2（ab+ac+bc）=1+0
(a+b+c)^2=1
所以a+b+c=1或-1

综上所述a+b+c=0或1或-1

a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=-3
a(1/b+1/c)+1+b(1/c+1/a)+1+c(1/a+1/b)+1=-3+3
a(1/a+1/b+1/c)+b(1/a+1/b+1/c)+c(1/a+1/b+1/c)=0
(a+b+c)*(1/a+1/b+1/c)=0

a+b+c=0

或1/a+1/b+1/c=0
(bc+ac+ab)/(abc)=0
ab+ac+bc=0

a^2+b^2+c^2=1
a^2+b^2+c^2+2ab+2ac+2bc=1+0
(a+b+c)^2=1
a+b+c=1或-1

综上所述a+b+c=0或1或-1