11.1. Find all values of the parameter $a$ for which the equation $\left|x^{3}+1\right|=a(x+1)$ has three roots.
Answer: $3 / 40$ for all $x$, the equation can be written in the form $|x+1|\left(x^{2}-x+1\right)=a(x+1)$. The root $x=-1$ exists for any $a$. For $x \neq-1$, reduce the equation by $(x+1)$. We get the equation $x^{2}-x+1=a \cdot \operatorname{sign}(x+1)$, where $\operatorname{sign}($ ) denotes the sign of the number (in this case, it is +1 or -1, respectively, for $x>-1$ or $x-1$, the quadratic equation cannot have more than one root in the domain $x-1$ (considering the positivity of the discriminant, i.e., the radicand, as well as the parameter $a$ ). Then $\sqrt{4 a-3} \cdot<3$ and, therefore, $0<4 a-3<9 \Leftrightarrow 3 / 4<a<3$. Comment. The problem allows for other (graphical) solutions.