Answer
$1,2, \frac{-1\pm i\sqrt {3}}{2}, -1\pm i\sqrt 3$
Work Step by Step
$F(x)=x^6-9x^3+8=(x^3-1)(x^3-8)=(x-1)(x^2+x+1)(x-2)(x^2+2x+4)$. Let $F(x)=0$, we get:
1. $x=1,2$
2. $x=\frac{-1\pm\sqrt {1-4}}{2}=\frac{-1\pm i\sqrt {3}}{2}$
3. $x=\frac{-2\pm\sqrt {4-4(4)}}{2}=-1\pm i\sqrt 3$