Simplify

Nevermind, I got it right, I just forgot about the 2...

$\left( \frac{4}{x} \right)^{-1} \left( \frac{3x^3}{x} \right)^2 \left( \frac{1}{2x} \right)^{-3} = \frac{x}{4} \cdot (3x^2)^2 \cdot (2x)^3 = \frac{x}{4} \cdot 9x^4 \cdot 8x^3 = \boxed{18x^8}$.