1. Let positive numbers $x, y, z$ satisfy
$$
\frac{1}{x^{3}}=\frac{8}{y^{3}}=\frac{27}{z^{3}}=\frac{k}{(x+y+z)^{3}} \text {. }
$$

Then $k=$ $\qquad$
$$
\begin{array}{l}
\sqrt[3]{k}=\frac{x+y+z}{x}=\frac{2(x+y+z)}{y} \\
=\frac{3(x+y+z)}{z}=\frac{6(x+y+z)}{x+y+z}=6 .
\end{array}
$$

Therefore, $k=216$.