1. (40 points) Find all real roots of the equation
$$
x^{2}-x+1=\left(x^{2}+x+1\right)\left(x^{2}+2 x+4\right)
$$
1. The original equation can be transformed into
$$
x^{4}+3 x^{3}+6 x^{2}+7 x+3=0 \text {. }
$$

Let $x=-1$. Then $1-3+6-7+3=0$.
Therefore, $x+1$ is a factor.
$$
\begin{array}{l}
\text { Hence } x^{4}+3 x^{3}+6 x^{2}+7 x+3 \\
=(x+1)\left(x^{3}+2 x^{2}+4 x+3\right) \\
=(x+1)^{2}\left(x^{2}+x+3\right)=0 .
\end{array}
$$

Notice that $x^{2}+x+3=\left(x+\frac{1}{2}\right)^{2}+\frac{11}{4}>0$.
Thus, the only real root is -1.