A and B are square matrices of same order such that `A^5=B^5` and `A^2B^3=B^2A^3` . If `A^2-B^2` invertible then `det(A^3+B^3)=?`

Answer

To solve the problem, we need to find the value of \(\det(A^3 + B^3)\) given the conditions \(A^5 = B^5\), \(A^2B^3 = B^2A^3\), and that \(A^2 - B^2\) is invertible.

### Step-by-step Solution:

1. **Understanding the Given Conditions**:
- We have two square matrices \(A\) and \(B\) of the same order.
- The first condition \(A^5 = B^5\) implies that the fifth powers of both matrices are equal.
- The second condition \(A^2B^3 = B^2A^3\) indicates a specific relationship between the matrices when raised to certain powers.
- The third condition states that \(A^2 - B^2\) is invertible, which means \(\det(A^2 - B^2) \neq 0\).

2. **Using the Identity for Cubes**:
- We can use the identity for the sum of cubes:
\[ A^3 + B^3 = (A + B)(A^2 - AB + B^2) \]
- To find \(\det(A^3 + B^3)\), we can express it as:
\[ \det(A^3 + B^3) = \det((A + B)(A^2 - AB + B^2)) \]

3. **Applying Determinant Properties**:
- Using the property of determinants, we have:
\[ \det(A^3 + B^3) = \det(A + B) \cdot \det(A^2 - AB + B^2) \]

4. **Relating the Given Conditions**:
- From the condition \(A^5 = B^5\), we can infer that \(A\) and \(B\) might have some common eigenvalues or roots.
- From \(A^2B^3 = B^2A^3\), we can manipulate this to find relationships between \(A\) and \(B\).

5. **Using the Invertibility Condition**:
- Since \(A^2 - B^2\) is invertible, we can express it as:
\[ A^2 - B^2 = (A - B)(A + B) \]
- The invertibility of \(A^2 - B^2\) implies that neither \(A - B\) nor \(A + B\) can be singular.

6. **Conclusion**:
- By multiplying \(A^3 + B^3\) by \(A^2 - B^2\), we can derive that:
\[ \det(A^3 + B^3)(\det(A^2 - B^2)) = 0 \]
- Since \(\det(A^2 - B^2) \neq 0\), it follows that:
\[ \det(A^3 + B^3) = 0 \]

### Final Answer:
\[ \det(A^3 + B^3) = 0 \]