A and B are different matrices of order n satisfying `A^(3)=B^(3)` and `A^(2)B=B^(2)A`. If det. `(A-B) ne 0`, then find the value of det. `(A^(2)+B^(2))`.

Explanation

To solve the problem, we need to find the value of \( \det(A^2 + B^2) \) given the conditions \( A^3 = B^3 \) and \( A^2B = B^2A \), along with the fact that \( \det(A - B) \neq 0 \).

### Step-by-Step Solution:

1. **Given Conditions**:
We start with the equations:
\[ A^3 = B^3 \quad \text{(1)} \]
\[ A^2B = B^2A \quad \text{(2)} \]

2. **Using the Identity**:
We can use the identity:
\[ A^2 + B^2 = (A - B)(A + B) + AB + BA \]
Rearranging gives:
\[ A^2 + B^2 = (A - B)(A + B) + AB + BA \]

3. **Substituting the Given Conditions**:
From equation (1), we know that:
\[ A^3 - B^3 = (A - B)(A^2 + AB + B^2) = 0 \]
Since \( A^3 = B^3 \), we can factor this as:
\[ (A - B)(A^2 + AB + B^2) = 0 \]
Given that \( \det(A - B) \neq 0 \), it implies that \( A - B \) is invertible, and thus:
\[ A^2 + AB + B^2 = 0 \quad \text{(3)} \]

4. **Rearranging Equation (3)**:
Rearranging equation (3) gives:
\[ A^2 + B^2 = -AB \]

5. **Finding the Determinant**:
Now, we need to find \( \det(A^2 + B^2) \):
\[ \det(A^2 + B^2) = \det(-AB) = (-1)^n \det(A) \det(B) \]

6. **Using the Properties of Determinants**:
Since \( A^2 + B^2 = -AB \) and \( A \) and \( B \) are different matrices, we can conclude that:
\[ \det(A^2 + B^2) = 0 \]

### Final Result:
Thus, the value of \( \det(A^2 + B^2) \) is:
\[ \boxed{0} \]