38.2. Solve the equation

$$
\left(x^{2}+x+1\right)\left(3-x-x^{2}\right)=3
$$

$$
\text { (8-10 grades) }
$$
38.2. Setting $t=x^{2}+x+1$, we get the equation:

$$
t(4-t)=3 \text { or } t^{2}-4 t+3=0 .
$$

From this, $t_{1}=1, t_{2}=3$. Now it remains to solve the equations:

$$
x^{2}+x+1=1 \quad \text { and } \quad x^{2}+x+1=3 \text {. }
$$

We obtain four roots of the given equation: $x_{1}=0, x_{2}=-1, x_{3}=1$, $x_{4}=-2$.