Find all real roots of the equation
$$
x^{6}-x^{3}+1=\left(x^{6}+x^{3}+1\right)\left(x^{2}+2 x+4\right)
$$
Solve:
$$
\begin{array}{l}
x^{6}-x^{3}+1=\left(x^{6}+x^{3}+1\right)\left(x^{2}+2 x+4\right) \\
\Rightarrow \frac{x^{6}-x^{3}+1}{x^{6}+x^{3}+1}=x^{2}+2 x+4 \\
\Rightarrow \frac{x^{6}-x^{3}+1}{x^{6}+x^{3}+1}-3=x^{2}+2 x+1 \\
\Rightarrow \frac{-2\left(x^{3}+1\right)^{2}}{x^{6}+x^{3}+1}=(x+1)^{2} \\
\Rightarrow(x+1)^{2}\left[1+\frac{2\left(x^{2}-x+1\right)^{2}}{x^{6}+x^{3}+1}\right]=0 .
\end{array}
$$

Notice that, $x^{6}+x^{3}+1=\left(x^{3}+\frac{1}{2}\right)^{2}+\frac{3}{4}>0$.
Therefore, the original equation has a unique real root -1.