Let $a=\sqrt[3]{n}$ and $b=$ Number formed from the last three digits of $n$ Then $b\in\mathbb{Z}^+\cup\{0\}$ Then $a=\frac{n-b}{1000}\in\mathbb{Z}^+$ $0\leq b\leq 999\implies 0\leq a^3-1000a\leq 999$ $0\leq a^3-1000a\implies 0\leq a^2-1000\implies 31^2<1000\leq a^2\implies 32\leq a$ $a^3-1000a\leq 999\implies a^3+1\leq 1000(a+1)\implies a^2-a+1\leq 1000 $ $\implies (2a-1)^2\leq 3997< 64^2\implies a\leq 32$ So $32\leq a\leq 32\implies a=32\implies n=32^3=32768$ and clearly this is a solution. So $n=32^{3}=32768$ is the only solution.