1A. Solve the equation in $\mathbb{R}$

$$
\left(x^{2}-3\right)^{3}-(4 x+6)^{3}+6^{3}=18(4 x+6)\left(3-x^{2}\right)
$$
Solution. With the substitution:

$$
a=x^{2}-3, b=-(4 x+6), c=6
$$

the equation takes the form

$$
\begin{gathered}
a^{3}+b^{3}+c^{3}-3 a b c=0 \Leftrightarrow(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-a c\right)=0 \\
\quad \Leftrightarrow(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(a-c)^{2}\right]=0 \\
\Leftrightarrow a+b+c=0 \text { or } a-b=a-c=b-c=0 \\
\quad \Leftrightarrow x^{2}-4 x-3=0 \text { or }\left\{\begin{array}{l}
x^{2}+4 x+3=0 \\
x^{2}-9=0 \\
-4 x-12=0
\end{array}\right. \\
\quad \Leftrightarrow x_{1}=2+\sqrt{7}, x_{2}=2-\sqrt{7}, x_{3}=-3 .
\end{gathered}
$$