因为a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=-3
所以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
∵a^2+b^2+c^2=1
(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=1+2(ab+bc+ca)
ab+bc+ca=[(a+b+c)^2-1]/2
∴(a+b+c)(ab+bc+ca)/abc=(a+b+c)*[(a+b+c)^2-1]/(2abc)=0
1:a+b+c=0
2:a+b+c=±1