AX方+BX+C=0,BX方+CX+A=0,CX方+CX方+AX+BX=0有一公根
(a+b+c)x^2+(a+b+c)x+(a+b+c)=0
(a+b+c)(x^2+x+1)=0
因为x^2+x+1=(x+1/2)^2+3/4>0
所以,a+b+c=0
(A三次方+B三次方+C三次方)除以ABC
=(a^3+b^3+c^3-3abc)/abc+3abc/abc
=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)/abc+3
=0+3
=3