10.3. The equation $P(x)=0$, where $P(x)=x^{2}+b x+c$, has a unique root, and the equation $P(P(P(x)))=0$ has exactly three distinct roots. Solve the equation $P(P(P(x)))=0$.
Answer: $1 ; 1+\sqrt{2} ; 1-\sqrt{2}$.

Solution. From the condition, it follows that $P(x)=(x-\alpha)^{2}$. The equation $P(P(P(x)))=0$ has the form

$$
\left(\left((x-\alpha)^{2}-\alpha\right)^{2}-\alpha\right)^{2}=0
$$

Thus $\left((x-\alpha)^{2}-\alpha\right)^{2}=\alpha$ and therefore $\alpha>0$. From the last equality, we get that $(x-\alpha)^{2}=\alpha \pm \sqrt{\alpha}$. Since the two resulting equations have exactly three solutions in total, one has two solutions, and the other has one solution. Therefore, $\alpha-\sqrt{\alpha}=0$, so $\alpha=1$. From this, we find the roots of the equation: $1 ; 1+\sqrt{2} ; 1-\sqrt{2}$.