If A is a square matrix such that `A^(2)= I`, then
`(A-I)^(3)+(A+I)^(3)-7A` is equal to

Options

A. A
B. `I-A`
C. I+A
D. 3A

Explanation

To solve the problem, we need to simplify the expression \((A-I)^{3} + (A+I)^{3} - 7A\) given that \(A^2 = I\).

### Step-by-Step Solution:

1. **Identify the expression**:
We start with the expression:
\[ (A-I)^{3} + (A+I)^{3} - 7A \]

2. **Use the identity for the sum of cubes**:
Recall the identity for the sum of cubes:
\[ x^3 + y^3 = (x+y)(x^2 - xy + y^2) \]
Let \(x = A - I\) and \(y = A + I\). Then:
\[ x + y = (A - I) + (A + I) = 2A \]
Now, we need to compute \(x^2 - xy + y^2\).

3. **Calculate \(x^2\)**:
\[ x^2 = (A - I)^{2} = A^2 - 2A + I = I - 2A + I = 2I - 2A \]

4. **Calculate \(y^2\)**:
\[ y^2 = (A + I)^{2} = A^2 + 2A + I = I + 2A + I = 2I + 2A \]

5. **Calculate \(xy\)**:
\[ xy = (A - I)(A + I) = A^2 - I^2 = I - I = 0 \]

6. **Combine the results**:
Now we can substitute back into the sum of cubes:
\[ x^2 - xy + y^2 = (2I - 2A) - 0 + (2I + 2A) = 4I \]

7. **Substituting back into the sum of cubes**:
Now substituting back into the sum of cubes:
\[ (A-I)^{3} + (A+I)^{3} = (2A)(4I) = 8A \]

8. **Final expression**:
Now we substitute this back into the original expression:
\[ 8A - 7A = A \]

### Final Answer:
Thus, the expression \((A-I)^{3} + (A+I)^{3} - 7A\) simplifies to:
\[ \boxed{A} \]