Roots of Resolvent of Cubic

Theorem
Let $P$ be the cubic equation:
 * $a x^3 + b x^2 + c x + d = 0$ with $a \ne 0$

Let $\map r P$ be the resolvent equation of $P$, given by:
 * $u^6 - 2 R u^3 - Q^3$

Let the roots of $P$ be $\alpha_1, \alpha_2, \alpha_3$.

Then the roots of $\map r P$ can be expressed as:

where $\omega = -\dfrac {-1 + \sqrt {-3} } 2$ is one of the primitive (complex) cube roots of $1$.