There exist infinitely many integers $q$ such that $3^{q-1}\cdot q\cdot p$ is cubic number？ Let $p$ is give postive integers, show that:there exist infinitely many integers $q$ such that $3^{q-1}\cdot q\cdot p$ is cubic number?


Alternatively, I considered the form $q=p^2\cdot l^3$?But How to prove there exist infinitely $l$ such $p^2\cdot l^3-1\equiv0\pmod 3?$I can't figure it out
Hint.

Assume first that $3\not\mid p$. 

Set $q=3k$. (If $3\mid p$ but $9\not \mid p$, set $q=3k+1$, while if $9\mid p$ but $27\not \mid p$ set $q=3k+2$, etc). Then 
$$
N=3^{q-1}qp=(3^{k})^3kq
$$
It thus suffices to show that there exist infinitely many $k$, such that $pk$ is a cubic number.
Hint.

Assume first that $3\not\mid p$. 

Set $q=3k$. (If $3\mid p$ but $9\not \mid p$, set $q=3k+1$, while if $9\mid p$ but $27\not \mid p$ set $q=3k+2$, etc). Then 
$$
N=3^{q-1}qp=(3^{k})^3kq
$$
It thus suffices to show that there exist infinitely many $k$, such that $pk$ is a cubic number.