Example $\mathbf{5}$ equation
$$
\begin{array}{l}
\left(x^{3}-3 x^{2}+x-2\right)\left(x^{3}-x^{2}-4 x+7\right)+ \\
6 x^{2}-15 x+18=0
\end{array}
$$

all distinct real roots are $\qquad$
Let $x^{3}-3 x^{2}+x-2=A$, $x^{3}-x^{2}-4 x+7=B$.
It is easy to see that $6 x^{2}-15 x+18=3(B-A)-9$. Therefore, the original equation can be rewritten as $A B+3(B-A)-9=0$, which simplifies to $(A+3)(B-3)=0$.
If $A+3=0$, then
$$
\begin{array}{l}
x^{3}-3 x^{2}+x+1=0 \\
\Rightarrow(x-1)\left(x^{2}-2 x-1\right)=0 \\
\Rightarrow x_{1}=1, x_{2}=1+\sqrt{2}, x_{3}=1-\sqrt{2} .
\end{array}
$$

If $B-3=0$, then
$$
\begin{array}{l}
x^{3}-x^{2}-4 x+4=0 \\
\Rightarrow(x-1)(x+2)(x-2)=0 \\
\Rightarrow x_{4}=1, x_{5}=-2, x_{6}=2 .
\end{array}
$$

Therefore, all distinct real roots of the original equation are $1, \pm 2, 1 \pm \sqrt{2}$.