Is it possible to factorize the cubic $$1 + 2\cos(\theta_1) x + 4i\sin(\theta_1) \cos(\theta_2) x^2 + x^3=0$$ without explicitly using the cubic equation? Given the coefficients, the equation will expand to $$1 + (e^{i\theta_1} + e^{-i\theta_1}) x + (e^{i\theta_1} - e^{-i\theta_1})(e^{i \theta_2} + e^{i \theta_2}) x^2 + x^3=0$$ but I am unsure of how best to proceed.
Note 1: I can already factor the analogous quadratic equation
$$ \begin{align} & 1 + 2x \cos(\theta) + x^2 = \\ & 1 + (e^{i\theta} + e^{-i \theta}) x + x^2 = \\ & (1 + e^{i \theta}x) (1 + e^{-i\theta}x) = \\ & (e^{i\theta} + x) (e^{-i \theta} + x) = 0 \end{align} $$
Note 2: I had also entertained using a Tschirhausen transformation with the hope of simplifying the form of the equation.