Prove the inequality $\{\sqrt[3]{n}\}>\frac1{3\sqrt[3]{n^2}}$

Question

Prove that the inequality $$\{\sqrt[3]{n}\}>\frac1{3\sqrt[3]{n^2}}$$ holds for every positive integer $n$ not equal to a cube of an integer.

My work so far:

$0\le\{\sqrt[3]{n}\}<1$

$\{\sqrt[3]{n}\}=\sqrt[3]{n}-\lfloor \sqrt[3]{n}\rfloor$. Let $\lfloor \sqrt[3]{n}\rfloor=k \in \mathbb Z$. Then $$k<\sqrt[3]{n}-\frac1{3\sqrt[3]{n^2}}$$

Answer 1

Let $k=\lfloor n^{1/3} \rfloor$. Then $k^3 \leq n-1$ so that $(n^{1/3}-k)(n^{2/3}+k^2+kn^{1/3}) \geq 1$. Hope you can take it from here.