1. Given
$$
x \sqrt{x^{2}+3 x+18}-x \sqrt{x^{2}-6 x+18}=1 \text {. }
$$

then the value of $2 x \sqrt{x^{2}-6 x+18}-9 x^{3}$ is
$$
\text { II, 1. }-1 \text {. }
$$

Notice that
$$
\begin{array}{l}
x \sqrt{x^{2}+3 x+18}+x \sqrt{x^{2}-6 x+18} \\
=\frac{x^{2}\left(x^{2}+3 x+18\right)-x^{2}\left(x^{2}-6 x+18\right)}{x \sqrt{x^{2}+3 x+18}-x \sqrt{x^{2}-6 x+18}} \\
=\frac{x^{2} \cdot 9 x}{1}=9 x^{3} .
\end{array}
$$

Subtracting the above equation from the given equation, we get
$$
2 x \sqrt{x^{2}-6 x+18}=9 x^{3}-1 \text {. }
$$

Therefore, $2 x \sqrt{x^{2}-6 x+18}-9 x^{3}=-1$.