To solve the problem, we need to:

1. Find eigenvalues and eigenvectors of matrix $A$: This will allow us to construct $P$ (matrix of eigenvectors) and $D$ (diagonal matrix with eigenvalues).

2. Express $A^6$: Using the diagonalization formula $A = P D P^{-1}$, we find $A^6 = P D^6 P^{-1}$.

Let's compute this step by step.

Step 1: Matrix $A$

$A = \begin{bmatrix} 27 & 12 \\ -56 & -25 \end{bmatrix}$

Step 2: Find eigenvalues ($\lambda$):

Solve the characteristic equation $\text{det}(A - \lambda I) = 0$: $\text{det} \begin{bmatrix} 27 - \lambda & 12 \\ -56 & -25 - \lambda \end{bmatrix} = 0$ $(27 - \lambda)(-25 - \lambda) - (12)(-56) = 0$ Simplify: $\lambda^2 - 2\lambda - 1071 = 0$ Solve the quadratic equation: $\lambda = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1071)}}{2(1)}$ $\lambda = \frac{2 \pm \sqrt{4288}}{2} = \frac{2 \pm 2\sqrt{1072}}{2} = 1 \pm \sqrt{1072}$

Thus, the eigenvalues are: $\lambda_1 = 1 + \sqrt{1072}, \quad \lambda_2 = 1 - \sqrt{1072}.$

Step 3: Find eigenvectors:

For each eigenvalue $\lambda$, solve $(A - \lambda I)x = 0$ to find eigenvectors.

For $\lambda_1 = 1 + \sqrt{1072}$:

Solve $\begin{bmatrix} 27 - (1 + \sqrt{1072}) & 12 \\ -56 & -25 - (1 + \sqrt{1072}) \end{bmatrix}x = 0$.

For $\lambda_2 = 1 - \sqrt{1072}$:

Similarly, solve $\begin{bmatrix} 27 - (1 - \sqrt{1072}) & 12 \\ -56 & -25 - (1 - \sqrt{1072}) \end{bmatrix}x = 0$.

The eigenvectors form the columns of $P$.

Step 4: Diagonal matrix $D$: