1.已知a+b+c=0,d+e+f=0,求证a3+b3+c3/d3+e3+f3=abc/def 2.若x2+x=1=0,

1.证：∵a+b+c=0,∴(a+b)^3=-c^3,∴(a+b)=-c.
a^3+b^3+c^3=-3ab(a+b)=3abc.
同理：d^3+e^3+f^3=3def
∴(a^3+b^3+c^3)/(d^3+e^3+f^3)=3abc/3def=abc/def.
证毕.
2.若x^2+x=1=0
改为：x^2+x+1=0.
∵x^3-1=(x-1)(x^2+x+1)=0,x^3-1=0,∴x^3=1.
x^6+1/x^6=(x^3)^2+1/(x^3)^2=1+1=2.