66th Belarusian Mathematical Olympiad

Set $x=2-\frac{4}{\alpha }$ in the left hand side of the given equation: $$\begin{array}{rl}& {\left(2-\frac{4}{\alpha }\right)}^3-12\left(2-\frac{4}{\alpha }\right)+8=8\left({\left(1-\frac{2}{\alpha }\right)}^3-3\left(1-\frac{2}{\alpha }\right)+1\right)=\\ & =8\left(1-\frac{6}{\alpha }+\frac{12}{{\alpha }^2}-\frac{8}{{\alpha }^3}-3+\frac{6}{\alpha }+1\right)=8\left(-1+\frac{12}{{\alpha }^2}-\frac{8}{{\alpha }^3}\right)=\\ & =-\frac{8}{{\alpha }^3}\left({\alpha }^3-12\alpha +8\right)=0\end{array}$$ since by the problem condition ${\alpha }^3-12\alpha +8=0$. Therefore, the number $2-\frac{4}{\alpha }$ is a root of the given equation.