Note that \(\displaystyle 120=8\cdot 3\cdot 5\) and \(\displaystyle 9n^5-5n^3-4n=(n-1)n(n+1)(4+9n^2)\). It suffices to show that \(\displaystyle 8, 3, \text{ and } 5\) divide \(\displaystyle (n-1)n(n+1)(4+9n^2)\) for any \(\displaystyle n\).

Can you take it from here?