Show that if $$Z = 2(-H^\frac{1}{2})\cos(\theta),$$ where $Z$ is a root of the cubic equation $$Z^3 + 3HZ+G=0,$$ then $$\cos(3\theta) = \sqrt{-G^2/4H^3}.$$ Deduce that if the $H < 0$, and $G^2+4H^3 < 0$ the cubic equation has three real roots. By using the expression $\cos(3\theta) = \frac{1}{2}(e^{3j\theta}+e^{-3j\theta})$, solve the equation when $G^2+4H^3> 0$, treating seperately the cases $H < 0$ and $H > 0.$
Part of my solution Part 2 of my solution
Now for the other parts of the questions, I don't know how to go about it