已知非零实数a.b.c满足a^2+b^2+c^2=1,且a(1/b=1/c)+b(1/c+1/a)+c(1/a+1/b)=-3,求a+b+c的值

∵a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=-3
∴a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3=0
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)(ab+bc+ca)/abc=0
若a+b+c=0,则问题得解.
若ab+bc+ca=0,又因为(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)
故(a+b+c)^2=1+0=1
a+b+c=1或-1