Roots of polynomial of degree 6

Question

I'm struggling to find the complex roots of $x^6-9x^3+8 = 0$. I've managed to find the real roots (1 and 2) by letting a variable, say $α = x^3$ and substituting where relevant, leading to a quadratic equation which I subsequently solved by factorization. I know this method is not at all helpful in finding complex roots though. :(

I would appreciate it if you could point me to the simplest way of finding the complex roots of this specific equation.

Thank you.

Answer 1

Yes, what you have done is a very good start for the full root calculation. You now need to solve $\alpha^3=1$ and $\alpha^3=8$. The roots of the second equation are twice the roots of the first.

To solve $\alpha^3=1$, note that $\alpha^3-1=(\alpha-1)(\alpha^2+\alpha+1)$.

Answer 2

The method which you mentioned is helpful. In fact with $\alpha=x^3$ we have $$\alpha^2-9\alpha+8=0$$ so the roots are $$\alpha_1=1\quad \alpha_2=8$$ now $$x^3=1\iff x=\exp\left(\frac{2ik\pi}{3}\right),\quad k=0,1,2$$ and $$x^3=8\iff x=2\exp\left(\frac{2ik\pi}{3}\right),\quad k=0,1,2$$

Answer 3

Solving $x^3=1$ gives $x=1$ and the other roots satisfy x^2+x+1=0. For these you may use the quadratic formula $x=\frac{-1\pm \sqrt{-3}}{2}$.

Solving $x^3=8$ gives x=2 and the other roots satisfy x^2+2x+4=0. Use the quadratic formula $x=2\frac{-1\pm \sqrt{-3}}{2}=-1\pm \sqrt{-3}$.