Complex number - solve for all values of z

how do we do something like this question

z^6 + 8z^3 - 3=O

There is no universal solution for the sixth degree polynomial equation. However, for this particular equation you can substitute a=z^3

so you will end up with quadratic equation a^2+8a-3=0

Solve equation to find roots -4+/-sqrt(19), or

a = 0.3589
a = -8.359.

Then you will have to take third root from both of these numbers to get two answers


Edit: I noticed that the question was posted in the “complex numbers”, that means we have to calculate all six roots of the equation, including imaginary roots (the amount of roots of a polynomial equation is always equal to the degree of the equation, however, some roots can be imaginary).

To find all roots we will have to apply De Moivre’s theorem: