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 expression \((A-I)^{3} + (A+I)^{3} - 7A\) given that \(A^2 = I\), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ (A-I)^{3} + (A+I)^{3} - 7A \]

### Step 2: Apply the sum of cubes formula
Recall the formula 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 calculate \(x^2 - xy + y^2\):
- \(x^2 = (A - I)^2 = A^2 - 2A + I = I - 2A + I = 2I - 2A\)
- \(y^2 = (A + I)^2 = A^2 + 2A + I = I + 2A + I = 2I + 2A\)
- \(xy = (A - I)(A + I) = A^2 - I^2 = I - I = 0\)

Now substituting these into the sum of cubes:
\[ (A-I)^{3} + (A+I)^{3} = (2A)((2I - 2A) - 0 + (2I + 2A)) = (2A)(4I) = 8A \]

### Step 3: Substitute back into the expression
Now we substitute back into the original expression:
\[ 8A - 7A = A \]

### Final Result
Thus, the value of the expression \((A-I)^{3} + (A+I)^{3} - 7A\) is:
\[ \boxed{A} \]