因为a+b+c=0
所以(a+b+c)^3 = a^3+b^3+c^3+3a^2b+3a^2c+3b^2c+3c^2a+3c^2b+6abc=0
abc不等于0,同时除以abc得：
a^2/bc +b^2/ac +c^2/ab +3a/c+3a/b+3b/c+3b/a+3c/b+3c/a+6=0
即:a^2/bc +b^2/ac +c^2/ab +3(b+c)/a +3(a+b)/c +3(a+c)/b = -6
因为a+b+c=0,所a+b=-c,a+c=-b,b+c=-a,代入上式
a^2/bc +b^2/ac +c^2/ab +3(-a)/a+3(-c)/c+3(-b)/b =-6
所以a^2/bc +b^2/ac +c^2/ab =-6 + 9 =3