Suppose A is square matrix such that `A^(3) =I ` then `(A+I)^(3) +(A-I)^(3)-6A` equals

Options

A. I
B. 2I
C. A
D. 3A

Answer

To solve the problem, we need to evaluate the expression \((A + I)^3 + (A - I)^3 - 6A\) given that \(A^3 = I\).

### Step-by-Step Solution:

1. **Expand \((A + I)^3\)**:
Using the binomial expansion:
\[ (A + I)^3 = A^3 + 3A^2I + 3AI^2 + I^3 \]
Since \(I^2 = I\) and \(I^3 = I\), we can simplify this to:
\[ (A + I)^3 = A^3 + 3A^2 + 3A + I \]

2. **Expand \((A - I)^3\)**:
Similarly, using the binomial expansion:
\[ (A - I)^3 = A^3 - 3A^2I + 3AI^2 - I^3 \]
This simplifies to:
\[ (A - I)^3 = A^3 - 3A^2 + 3A - I \]

3. **Combine the two expansions**:
Now, we add the two results:
\[ (A + I)^3 + (A - I)^3 = (A^3 + 3A^2 + 3A + I) + (A^3 - 3A^2 + 3A - I) \]
Simplifying this gives:
\[ 2A^3 + 6A \]

4. **Subtract \(6A\)**:
Now we substitute this into the original expression:
\[ (A + I)^3 + (A - I)^3 - 6A = (2A^3 + 6A) - 6A \]
This simplifies to:
\[ 2A^3 \]

5. **Substitute \(A^3 = I\)**:
Since we know \(A^3 = I\), we substitute this into the equation:
\[ 2A^3 = 2I \]

### Final Answer:
Thus, the final result is:
\[ \boxed{2I} \]