If \(p \gt \sqrt[3]{n} \), then:
$$\begin{gather} p \gt \sqrt[3]{n} \\ 1/p \lt \frac{1}{\sqrt[3]{n}} \\ n/p \lt \frac{n}{\sqrt[3]{n}} \end{gather} $$
Since \(n = (\sqrt[3]{n})(\sqrt[3]{n})(\sqrt[3]{n})\):
$$\begin{gather} n/p \lt \frac{(\sqrt[3]{n})(\sqrt[3]{n})(\sqrt[3]{n})}{\sqrt[3]{n}} \\ n/p \lt \sqrt[3]{n}^2 \\ \sqrt{n/p} \lt \sqrt[3]{n} \end{gather} $$
For any integer \(x \gt 1\), \(x\) must have a smallest prime factor less than \(\sqrt{x}\).
Assume that \(n/p\) is composite. For the integer \(n/p\), there must be a smallest prime factor less than \(\sqrt{n/p}\), call this \(q\). According to the above, \(q \lt \sqrt[3]{n}\).
Since \(q\) is a factor of \(n/p\), then it is also a factor of \(n\). However, the smallest prime factor of \(n\) exceeds \(\sqrt[3]{n}\). If \(q \lt \sqrt[3]{n}\), then we have a contradiction.
This means that \(n/p\) cannot be composite.