Determine all real numbers $x$ for which $x^{4}-3 x^{3}+x^{2}-3 x=0$.
Factoring, we obtain the following equivalent equations:

$$
\begin{aligned}
x^{4}-3 x^{3}+x^{2}-3 x & =0 \\
x\left(x^{3}-3 x^{2}+x-3\right) & =0 \\
x\left[x^{2}(x-3)+(x-3)\right] & =0 \\
x(x-3)\left[x^{2}+1\right] & =0
\end{aligned}
$$

Therefore, $x=0$ or $x-3=0$ (which gives $x=3$ ) or $x^{2}+1=0$ (which has no real solutions). Therefore, $x=0$ or $x=3$.

ANSWER: $x=0,3$

#