Let `A`, `B` are square matrices of same order satisfying `AB=A` and `BA=B` then `(A^(2010)+B^(2010))^(2011)` equals.

Options

A. `A+B`
B. `2010(A+B)`
C. `2011(A+B)`
D. `2^(2011)(A+B)`

Explanation

To solve the problem, we need to find the expression \((A^{2010} + B^{2010})^{2011}\) given that \(AB = A\) and \(BA = B\).

### Step-by-Step Solution:

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

2. **Multiply Equation (1) by \(A\)**:
From equation (1), multiply both sides by \(A\) on the right:
\[ A(BA) = AA \implies A(B) = A^2 \implies A = A^2 \]
This implies that \(A\) is idempotent.

3. **Multiply Equation (2) by \(B\)**:
From equation (2), multiply both sides by \(B\) on the right:
\[ B(AB) = BB \implies B(A) = B^2 \implies B = B^2 \]
This implies that \(B\) is also idempotent.

4. **Generalizing the Idempotent Property**:
Since \(A = A^2\), we can generalize this to:
\[ A^n = A \quad \text{for any integer } n \geq 1 \]
Similarly, for \(B\):
\[ B^n = B \quad \text{for any integer } n \geq 1 \]

5. **Finding \(A^{2010}\) and \(B^{2010}\)**:
Using the idempotent property:
\[ A^{2010} = A \quad \text{and} \quad B^{2010} = B \]

6. **Substituting Back into the Expression**:
Now substituting these results back into the original expression:
\[ A^{2010} + B^{2010} = A + B \]

7. **Final Expression**:
We now raise the result to the power of 2011:
\[ (A^{2010} + B^{2010})^{2011} = (A + B)^{2011} \]

8. **Using the Binomial Expansion**:
To find \((A + B)^{2011}\), we can use the binomial theorem. However, we can also find a simpler relation. We can check:
\[ (A + B)^2 = A^2 + B^2 + AB + BA = A + B + A + B = 2(A + B) \]
This implies:
\[ (A + B)^n = 2^{n-1}(A + B) \quad \text{for } n \geq 1 \]

9. **Final Calculation**:
For \(n = 2011\):
\[ (A + B)^{2011} = 2^{2010}(A + B) \]

### Conclusion:
Thus, the final answer is:
\[ (A^{2010} + B^{2010})^{2011} = 2^{2010}(A + B) \]