1. Given $x$ is a real number,
$$
\sqrt{x^{3}+2020}-\sqrt{2030-x^{3}}=54 \text {. }
$$

Then $28 \sqrt{x^{3}+2020}+27 \sqrt{2030-x^{3}}=$
II. 1.2007.
Let $x^{3}+2020=a, 2030-x^{3}=b$. Then $a+b=4050$.
From the problem, we have
$$
\begin{array}{l}
\sqrt{a}-\sqrt{b}=54 \\
\Rightarrow a+b-2 \sqrt{a b}=54^{2}=2916 \\
\Rightarrow 2 \sqrt{a b}=(a+b)-2916 \\
\quad=4050-2916=1134 \\
\Rightarrow(\sqrt{a}+\sqrt{b})^{2}=(a+b)+2 \sqrt{a b} \\
\quad=4050+1134=5184 \\
\Rightarrow \sqrt{a}+\sqrt{b}=72 .
\end{array}
$$

From equations (1) and (2), we get $\sqrt{a}=63, \sqrt{b}=9$.
Therefore, $28 \sqrt{x^{3}+2020}+27 \sqrt{2030-x^{3}}$ $=2007$.