(6j^-2)^-3

$$\\(6j^{-2})^{-3}\\ =6^{-3}\;\;j^{-2*-3}\\\\ =\frac{6^{-3}\;\;j^{6}}{1}\\\\ =\frac{j^{6}}{6^3}\\\\ =\frac{j^{6}}{216}\\\\ $Now if j is an ordinary pronumeral then the question is finished$\\ $but if j is the imaginary number $ \sqrt{-1} \qquad then\\\\ (\sqrt{-1})^6=((\sqrt{-1})^2)^3=(-1)^3=-1\\\\ $ so the final answer would be $ \frac{-1}{216}$$