$1 \cdot 146$ Simplify $\left(\frac{1 \cdot 2 \cdot 4+2 \cdot 4 \cdot 8+\cdots+n \cdot 2 n \cdot 4 n}{1 \cdot 3 \cdot 9+2 \cdot 6 \cdot 18+\cdots+n \cdot 3 n \cdot 9 n}\right)^{\frac{1}{3}}$.
[Solution] Since for $1 \leqslant k \leqslant n$ we have
$$
\begin{array}{l}
k \cdot 2 k \cdot 4 k=8 k^{3}, \\
k \cdot 3 k \cdot 9 k=27 k^{3} .
\end{array}
$$
$$
\text { Therefore, the original expression } \begin{aligned}
& =\left[\frac{8\left(1^{3}+2^{3}+\cdots+n^{3}\right)}{27\left(1^{3}+2^{3}+\cdots+n^{3}\right)}\right]^{\frac{1}{3}} \\
& =\frac{2}{3} .
\end{aligned}
$$