设三个关于x的一元二次方程的公共实数根为t,
则at2+bt+c=0①,bt2+ct+a=0②,ct2+at+b=0③,
①+②+③得(a+b+c)t2+(a+b+c)t+(a+b+c)=0,
∴(a+b+c)(t2+t+1)=0,
而t2+t+1=(t+[1/2])2+[3/4],
∵(t+[1/2])2≥0,
∴t2+t+1>0,
∴a+b+c=0,
∴a+b=-c,
原式=
a3+b3+c3
abc
=
(a+b)(a2−ab+b2)+c3
abc
=
−c(a2−ab+b2)+c3
abc
=
c2−(a2−ab+b2)
ab
=
c2−[(a+b) 2−3ab]
ab
=
c2−c2+3ab
ab
=3.