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.
Best Answer
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 $\alpha^3-1=(\alpha-1)(\alpha^2+\alpha+1)$.