By Cayley–Hamilton, $A$ satisfies its characteristic polynomial ${\lambda }^2-\mathrm{tr}(A)\lambda +\det (A)=0$. Matching $A^2-4A+2I=0$ gives $\mathrm{tr}(A)=4$, so $\alpha =2$. Matching $B^2-3B+I=0$ gives $\mathrm{tr}(B)=3$, so $\beta =2$.

Use the relations to reduce $A^3$ and $B^3$ to linear combinations of $A,I$ and $B,I$:

$A^3=A\cdot A^2=A(4A-2I)=4A^2-2A=4(4A-2I)-2A=14A-8I=\left[\begin{array}{c}2028\\ 1420\end{array}\right]$.

$B^3=B(3B-I)=3B^2-B=3(3B-I)-B=8B-3I=\left[\begin{array}{c}58\\ 813\end{array}\right]$.

$A^3-B^3=\left[\begin{array}{c}1520\\ 67\end{array}\right]$, $\det =105-120=-15$.

For a $2\times 2$ matrix $M$, $\det (\mathrm{adj}M)=\det M$. So $(\det (\mathrm{adj}(A^3-B^3)){)}^2=(-15{)}^2=225$.