4. Equation
$$
\left(x^{3}-3 x^{2}+4 x-3\right)\left(x^{3}+2 x^{2}-x-1\right)+5 x^{2}-5 x+1=0
$$

The distinct real roots of the equation are ( ).
(A) $1,-1$
(B) $1,-1,-2$
(C) $1,-1,-2,2$
(D) None of the above
4. B.

Let $x^{3}-3 x^{2}+4 x-3=A, x^{3}+2 x^{2}-x-1=B$.
It is easy to see that $5 x^{2}-5 x+1=(B-A)-1$.
Therefore, the original equation can be transformed into
$$
A B+(B-A)-1=0 \Rightarrow(A+1)(B-1)=0 \text {. }
$$

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

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

Therefore, all distinct real roots of the original equation are $1, -1, -2$.