$\frac{1}{(3^3)^5}$ 

$8^{-x}$ 

Rewrite $16$ and $32$ each as a power of $2$.

$(2^4)^x\left(\frac{1}{2^5}\right)^{(2-x)}$ 

$2^{4x}\left(2^{-5}\right)^{(2-x)}$ 

$2^{4x}\cdot 2^{-10+5x}$