1. Given that $a$ and $b$ are distinct real numbers, and
$$
(\sqrt[3]{a}+\sqrt[3]{b})^{3}=a^{2} b^{2} \text {. }
$$

then $(3 a+1)(3 b+1)-3 a^{2} b^{2}=$
$-1.1$
From $(\sqrt[3]{a}+\sqrt[3]{b})^{3}=a^{2} b^{2} \Rightarrow \sqrt[3]{a}+\sqrt[3]{b}=\sqrt[3]{a^{2} b^{2}}$.
Let $x=\sqrt[3]{a}, y=\sqrt[3]{b}, z=\sqrt[3]{a^{2} b^{2}}$. Then $x+y=z$.
$$
\begin{array}{l}
\text { Hence }(3 a+1)(3 b+1)-3 a^{2} b^{2} \\
=3\left(a+b+3 a b-a^{2} b^{2}\right)+1 \\
=3\left(x^{3}+y^{3}+3 x y z-z^{3}\right)+1 . \\
\text { By } x^{3}+y^{3}+3 x y z-z^{3} \\
=x^{3}+y^{3}+3 x y(x+y)-(x+y)^{3}=0 \\
\Rightarrow(3 a+1)(3 b+1)-3 a^{2} b^{2}=1 .
\end{array}
$$