Belarusian Mathematical Olympiad

Set $x = 2 - \frac{4}{\alpha}$ in the left hand side of the given equation:
$$\begin{aligned} & \left(2 - \frac{4}{\alpha}\right)^3 - 12\left(2 - \frac{4}{\alpha}\right) + 8 = 8\left(\left(1 - \frac{2}{\alpha}\right)^3 - 3\left(1 - \frac{2}{\alpha}\right) + 1\right) = \\ & = 8\left(1 - \frac{6}{\alpha} + \frac{12}{\alpha^2} - \frac{8}{\alpha^3} - 3 + \frac{6}{\alpha} + 1\right) = 8\left(-1 + \frac{12}{\alpha^2} - \frac{8}{\alpha^3}\right) = \\ & = -\frac{8}{\alpha^3}\left(\alpha^3 - 12\alpha + 8\right) = 0 \end{aligned}$$
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.