Question 975579
x^3+1=0
This has 3 possible zeros.  If there are any complex zeros, there can only be 1 rational root, since the complex zeros are conjugate.  If there are two rational zeros, they have to be the same. 

(x+1) (x^2-x+1)=0, factoring a sum of cubes.

There is 1 rational root, , -1
There are two complex roots

x=(1/2)  1 +./- sqrt (1-4)=  (1/2)+/- (1/2) * i * (sqrt 3)

{{{graph(300,200,-10,10,-10,10,x^3+1)}}}