4. 228 ** Let $x, y, z \in \mathbf{R}^{+}, x+2 y+3 z=1$, find the minimum value of $\frac{16}{x^{3}}+\frac{81}{8 y^{3}}+\frac{1}{27 z^{3}}$.
By the AM-GM inequality, we have
$$
\begin{array}{l}
\frac{16}{x^{3}}+6^{4} x+6^{4} x+6^{4} x \geqslant 4 \sqrt[4]{6^{4} \cdot 6^{4} \cdot 6^{4} \cdot 16}=12^{3}, \\
\frac{81}{8 y^{3}}+6^{4} \cdot 2 y+6^{4} \cdot 2 y+6^{4} \cdot 2 y \geqslant 4 \sqrt[4]{6^{4} \cdot 6^{4} \cdot 6^{4} \cdot 81}=12 \cdot 6^{3}, \\
\frac{1}{27 z^{3}}+6^{4} \cdot 3 z+6^{4} \cdot 3 z+6^{4} \cdot 3 z \geqslant 4 \sqrt[4]{6^{12}}=4 \cdot 6^{3},
\end{array}
$$

Adding the above three inequalities, we get
$$
\frac{16}{x^{3}}+\frac{81}{8 y^{3}}+\frac{1}{27 z^{3}}+3 \cdot 6^{4}(x+2 y+3 z) \geqslant 4 \cdot 6^{3}(2+3+1),
$$

Thus, $\frac{16}{x^{3}}+\frac{81}{8 y^{3}}+\frac{1}{27 z^{3}} \geqslant 6^{1}$, with equality holding if and only if $x=\frac{1}{3}, y=\frac{1}{4}, z=\frac{1}{18}$. Therefore, the minimum value sought is $6^{4}=1296$.
Another solution: Let $a=x, b=2 y, c=3 z$, then $a+b+c=1$, and let $S=\frac{16}{x^{3}}+\frac{81}{8 y^{3}}+\frac{1}{27 z^{3}}$, so $S=\frac{16}{a^{3}}+\frac{81}{b^{3}}+\frac{1}{c^{3}}$. By the Cauchy-Schwarz inequality,
$$
\left(\frac{16}{a^{3}}+\frac{81}{b^{3}}+\frac{1}{c^{3}}\right)(a+b+c) \geqslant\left(\frac{4}{a}+\frac{9}{b}+\frac{1}{c}\right)^{2},
$$

and $\left(\frac{1}{a}+\frac{9}{b}+\frac{1}{c}\right)(a+b+c) \geqslant(2+3+1)^{2}=6^{2}$,
so $S \geqslant\left(\frac{4}{a}+\frac{9}{b}+\frac{1}{c}\right)^{2} \geqslant 6^{4}$. Equality holds when $a=\frac{1}{3}, b=\frac{1}{2}, c=\frac{1}{6}$, i.e., when $x=\frac{1}{3}, y=\frac{1}{4}, z=\frac{1}{18}$, $S$ attains its minimum value $6^{4}=1296$.