Let  x3 = u
Then x6 = u2




The discriminant is 



which is negative u has two complex roots.

So there are two complex values x3.

There are 3 complex cube roots of each of those two complex
roots for u.  Therefore, 

there are 6 complex roots.  <--answer

Edwin

Regarding your equation 

 = 0,             (0)

we can solve it explicitly in this way.

First, introduce new variable y = .
In terms of this new equations, the original equation takes the form

 = 0.

It is a quadratic equation. You can solve it by applying the quadratic formula

 =  =  = .

So, one root is  = . The other root is  = .

You can write these complex numbers (complex roots) in a trigonometric (polar) form:

 =  =  = .  The modulus is  =  =  = 1  and the argument is .

 =  =  = .  The modulus is  =  =  = 1  and the argument is .


Next, since  = , then x =  = .


    Now, applying the de Moivre's formula, you have THREE values for x, that all are the cubic roots of the complex number :

         =  =  = ,            (1)

         =  =  = ,     (2)

         =  =  = .     (3)


By the same way as the root  produces the three solutions ,  and , the root   produces three other solutions ,  and .

Namely, since  = , you have x =  = .

    Now, applying the de Moivre's formula again, you have THREE other values for x, that all are the cubic roots of the complex number :

         =  =  = ,            (4)

         =  =  = ,     (5)

         =  =  = .     (6)

These complex numbers, , , , ,  and , are all different and are all the roots of the original equation (0).

Probably, this explanation is too complicated for the high school students.

But it is how the university students who study "High Algebra" or/and "Abstract Algebra" solve the equations like (0) in complex domain.

Expressions (1) - (6) could be simplified further and written in rectangular coordinates.
I will not go into these details here.

The complex numbers (1) - (6) lie on the unit circle of the complex domain.