$1 \cdot 146$ 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
$$\begin{aligned}
k \cdot 2 k \cdot 4 k & =8 k^{3} \\
k \cdot 3 k \cdot 9 k & =27 k^{3}
\end{aligned}$$

Therefore, the original expression $=\left[\frac{8\left(1^{3}+2^{3}+\cdots+n^{3}\right)}{27\left(1^{3}+2^{3}+\cdots+n^{3}\right)}\right]^{\frac{1}{3}}$
$$=\frac{2}{3} .$$