Roots of a cubic equation When will the roots of a cubic polynomial $ax^3+bx^2+cx+d$ generate a cyclic group of order 3?
I literally mean: the cubic equation $ax^3+bx^2+cx+d$ will have three roots (possibly all three real, possibly one real and two complex), say $\zeta_1$, $\zeta_2$ and $\zeta_3$. Under what conditions is $\{\zeta_1, \zeta_2, \zeta_3\}$ isomorphic to $C_3$?
If a complex number has multiplicative order 3, then it is a primitive cubic root of unit. Hence, two of the roots are primitive cubic roots of unit and the other one is 1. In other words, the polynomial is $a(x^3-1)$.
One obvious condition to get C3 as the Galois group is that there are no simple permutations of the roots. So a pair of complex roots is out. You must have three real roots. Then there must be some algebraic condition between the roots which does not allow you to permutate them willy-nilly. An example of three roots with built-in structure would be cos(pi/7), cos(2pi/7), and cos(3pi/7). They are the roots of the cubic equation which comes from adding together conjucate roots of x^7-1=0. If they don't have a Galois group of C3 then I'll be very surprised.