The 71st question: For a given positive integer $n (n \geq 3)$ that has no square factors greater than 1, find $\sum_{k=1}^{(n-2)(n-1)}[\sqrt[3]{k n}]$.


The 71st question:
Solution: Let the set $S$ be the lattice points in the rectangle $[1,(n-2)(n-1)] \times[1, n-1]$, then $|S|=(n-2)(n-1)^{2}$.

Since $n$ has no square factors greater than 1, there are no points in set $S$ on the curve $y=\sqrt[3]{nx}$. The sum we are looking for in this problem is the number of points in $S$ below the curve. The number of points above the curve is $\sum_{k=1}^{n-1}\left[\frac{k^{3}}{n}\right]$ (because the inverse function of $y=\sqrt[3]{nx}$ is $y=\frac{x^{3}}{n}$), hence
$$
\begin{array}{l}
\sum_{k=1}^{(n-2)(n-1)}[\sqrt[3]{kn}]=|S|-\sum_{k=1}^{n-1}\left[\frac{k^{3}}{n}\right] \\
=(n-2)(n-1)^{2}-\sum_{k=1}^{n-1}\left(\frac{k^{3}}{n}-\left\{\frac{k^{3}}{n}\right\}\right) \\
=(n-2)(n-1)^{2}-\sum_{k=1}^{n-1} \frac{k^{3}}{n}+\sum_{k=1}^{n-1}\left\{\frac{k^{3}}{n}\right\} \\
=(n-2)(n-1)^{2}-\frac{n^{2}(n-1)^{2}}{4 n}+\sum_{k=1}^{n-1}\left(\frac{k^{3}}{n}\right\}
\end{array}
$$

Since for any $k \in\{1,2, \ldots, n-1\},\left\{\frac{k^{3}}{n}\right\} \neq 0$, and $\frac{k^{3}}{n}+\frac{(n-k)^{3}}{n} \in \mathbb{Z}$, it follows that $\left\{\frac{k^{3}}{n}\right\}+\left\{\frac{(n-k)^{3}}{n}\right\}=1$. Therefore, we know:
$$
\begin{array}{l}
2 \cdot \sum_{k=1}^{n-1}\left\{\frac{k^{3}}{n}\right\}=\sum_{k=1}^{n-1}\left(\left\{\frac{k^{3}}{n}\right\}+\left\{\frac{(n-k)^{3}}{n}\right\}\right)=n-1 \\
\Rightarrow \sum_{k=1}^{n-1}\left\{\frac{k^{3}}{n}\right\}=\frac{n-1}{2}
\end{array}
$$

Substituting (2) into (1), we get:
$$
\sum_{k=1}^{(n-2)(n-1)}[\sqrt[3]{kn}]=(n-2)(n-1)^{2}-\frac{n^{2}(n-1)^{2}}{4 n}+\frac{n-1}{2}=\frac{(n-2)(n-1)(3 n-5)}{4}
$$

In conclusion, $\sum_{k=1}^{(n-2)(n-1)}[\sqrt[3]{kn}]=\frac{(n-2)(n-1)(3 n-5)}{4}$.