This equation is a quadratic in z^{3}...

(z^{3} - 4)(z^{3} + 2)

Use De Moivre's theorem to express the roots in polar form...although you really don't need De Moivre's theorem since the cube roots are simply obtained using the 3 cube roots of 1.

Yes...you need to use De Moivre's Theorem on each factor.

However, a short cut is that the 3 solutions of z^{3}=R are R^{1/3} times 1, times ω, and times ω^{2}.

.....ω and ω^{2} are the 2 complex roots of 1.

Henry R.

Hi thank you for your time so does this mean that I will have to apply the polar form and de moivre on each bracketed terms? z^3=4 --------- find 3 roots using polar form and de moivre ;; z^3=-2 -------- find 3 roots using polar form and de moivre thank you.01/24/21