设公共根是m
则am^2+bm+c=0
bm^2+cm+a=0
cm^2+am+b=0
相加
(a+b+c)m^2+(a+b+c)m+(a+b+c)=0
(a+b+c)(m^2+m+1)=0
因为m^2+m+1=(m+1/2)^2+3/4>0,即不会等于0
所以a+b+c=0
a=-b-c
a^2/bc+b^2/ca+c^2/ab
=(a^3+b^3+c^3)/abc
=[(-b-c)^2+b^3+c^3]/(-b-c)bc
=(-b^3-3b^2c-3bc^2-c^3+b^3+c^3)/(-b-c)bc
=(-3b^2c-3bc^2)/(-b-c)bc
=-3bc(b+c)/[-bc(b+c)]
=3