I am trying to express
$\frac{-9}{4}\left({\sqrt[3] 2}\right)^4+18\left({\sqrt[3] 2}\right)$
in the form
$2^p \cdot 3^q$
I have expressed both surds as fractional indices and the first fraction with $-18$ in the numerator so the denominator becomes $2^3.$ This forms part of a correct answer for $p$ but I can't see how to form $q$ (which should be $3$). Can anyone suggest how I might approach this?