Task B-2.2. Solve the equation in the set $\mathbb{C}$:

$$
x^{6}+x^{3}=18\left(12-x^{3}\right)
$$
Solution. By rearranging we get:

$$
\begin{array}{lr}
x^{6}+19 x^{3}-216=0 \\
t=x^{3} \\
t^{2}+19 t-216=0 & \\
t_{1,2}=\frac{-19 \pm 35}{2} & t_{2}=-27 \\
t_{1}=8 & x^{3}=-27 \\
x^{3}=8 & x^{3}+27=0 \\
x^{3}-8=0 & (x+3)\left(x^{2}-3 x+9\right)=0 \\
(x-2)\left(x^{2}+2 x+4\right)=0 & x+3=0, x^{2}-3 x+9=0 \\
x-2=0, x^{2}+2 x+4=0 & x_{4}=-3, x_{5,6}=\frac{3 \pm 3 i \sqrt{3}}{2}
\end{array}
$$

For $x_{1}$ and $x_{4}(1 \text{ point}), x_{2,3}(1 \text{ point})$ and for $x_{5,6}(1 \text{ point})$.