9. 5. First note that for any real numbers $x, y$, we have $x^{2}-x y+y^{2} \geqslant|x y|$.
The equality holds if and only if $x=y$.
If $a b c \neq 0$, then dividing the two expressions yields
$$\begin{array}{l} \left(a^{2}-a b+b^{2}\right)\left(b^{2}-b c+c^{2}\right)\left(c^{2}-c a+a^{2}\right) \\ =a^{2} b^{2} c^{2}=|a b| \cdot|b c| \cdot|a c| . \end{array}$$

The left side of the equation above is positive for each parenthesis, and the corresponding absolute values on the right side are also positive, hence $a=b=c$.
Thus, $8 a^{3}=a^{3} \Rightarrow a=0$. Contradiction.