What is the fastest technique to find complex roots of a function?

Question

I have the following function: $$ay^6+by^3+c=0$$ It can be re-written in the quadratic form: $$ax^2+bx+c=0$$

In my case, $a=1, b=-7, c=-8$; then the quadratic equation equals $(x-8)(x+1)$, therefore, the real roots are $x_1=8$ and $x_2=-1$.

But what is the fastest or simplest way to find complex roots?

Considering the theorem of complex numbers' trigonometric form: $$z^n=r^n(cos(n\phi)+isin(n\phi)),$$ where $z$ is a complex number, $r$ is the radius and $n$ is the degree; there must be $n$ (in our case $n=6$) roots totally.

I really appreciate any help you can provide.

*These are the complex solutions below. How can I calculate them without using computer?

Complex roots

Answer 1

In your case, the equation in $x$ is quadratic and you found both solutions. So now, you must solve $y^3 = 8$ and $y^3 = -1$.

Start with $y = re^{ia}$ so $y^3 = r^3 e^{i (3a)}$ and find $r$ for both equations and then find the angles. Update your answer or comment here and I will be glad to help further if needed.