For some $$\alpha,\beta\epsilon R$$, let $$A=\begin{bmatrix}\alpha &  2 \\ 1 &  2 \end{bmatrix}\text{ and }B=\begin{bmatrix}1 &  1 \\1 &   \beta \end{bmatrix}$$ be such that $$A^{2}-4A+2I=B^2-3B+I=0$$. Then $$(det(adj(A^3-B^3)))^2$$ is equal to _______.

$$A = \begin{bmatrix} \alpha & 2 \\ 1 & 2 \end{bmatrix}$$, $$B = \begin{bmatrix} 1 & 1 \\ 1 & \beta \end{bmatrix}$$.

$$A^2 - 4A + 2I = 0$$ and $$B^2 - 3B + I = 0$$.

For A: by Cayley-Hamilton, $$A^2 - (\text{tr}A)A + (\det A)I = 0$$.

$$\text{tr}A = \alpha + 2$$, $$\det A = 2\alpha - 2$$.

$$\alpha + 2 = 4 \Rightarrow \alpha = 2$$. $$2\alpha - 2 = 2 \Rightarrow \alpha = 2$$. ✓

For B: $$\text{tr}B = 1 + \beta = 3 \Rightarrow \beta = 2$$. $$\det B = \beta - 1 = 1$$. ✓

$$A = \begin{bmatrix} 2 & 2 \\ 1 & 2 \end{bmatrix}$$, $$\det A = 2$$. $$B = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$$, $$\det B = 1$$.

Using $$A^2 = 4A - 2I$$: $$A^3 = 4A^2 - 2A = 4(4A-2I) - 2A = 14A - 8I$$.

$$\det(A^3) = (\det A)^3 = 8$$.

Using $$B^2 = 3B - I$$: $$B^3 = 3B^2 - B = 3(3B-I) - B = 8B - 3I$$.

$$\det(B^3) = (\det B)^3 = 1$$.

$$A^3 - B^3 = (14A - 8I) - (8B - 3I) = 14A - 8B - 5I$$.

$$= 14\begin{bmatrix}2&2\\1&2\end{bmatrix} - 8\begin{bmatrix}1&1\\1&2\end{bmatrix} - 5I = \begin{bmatrix}28-8-5 & 28-8 \\ 14-8 & 28-16-5\end{bmatrix} = \begin{bmatrix}15&20\\6&7\end{bmatrix}$$

$$\det(A^3-B^3) = 105 - 120 = -15$$.

For 2×2 matrix: $$\text{adj}(M)$$ has $$\det(\text{adj}(M)) = (\det M)^{n-1} = (\det M)^1 = \det M$$.

$$(\det(\text{adj}(A^3-B^3)))^2 = (-15)^2 = 225$$.

The answer is 225.