Question

How many complex roots does the equation below have? x^6+x^3+1=0

Answer 1

Here, x⁶ + x³ + 1 = 0
Now, Let, x³ = u
Then, x⁶ = u²

So, u² + u + 1 = 0
We know, b² - 4ac = 1² - 4(1)(1) = 1 - 4 = -3

which is negative u has two complex roots.
So there are two complex values x³.
Again, There are 3 complex cube roots of each of those two complex roots for u. 

In short, Your Answer would be 3*2 = 6 Complex roots

Hope this helps!