1. Given condition:

$AA^T=I$

Since $A$ is a square matrix, this implies $A$ is orthogonal, hence

$A^T=A^{-1}$

and also

$A^TA=I.$

2. We need to simplify

$\frac12 A\left[(A+A^T)^2+(A-A^T)^2\right].$

3. Expand each square:

$(A+A^T)^2 = A^2+AA^T+A^TA+(A^T)^2$

and

$(A-A^T)^2 = A^2-AA^T-A^TA+(A^T)^2.$

4. Add them:

$(A+A^T)^2+(A-A^T)^2$ $= \left(A^2+AA^T+A^TA+(A^T)^2\right)+\left(A^2-AA^T-A^TA+(A^T)^2\right)$ $= 2A^2+2(A^T)^2.$

So the given expression becomes

$\frac12 A\left[2A^2+2(A^T)^2\right] = A\left[A^2+(A^T)^2\right].$

5. Multiply by $A$:

$A\left[A^2+(A^T)^2\right]=A^3+A(A^T)^2.$

Now use

$A(A^T)^2=(AA^T)A^T=IA^T=A^T.$

Therefore,

$A^3+A(A^T)^2=A^3+A^T.$

6. Hence the expression is

$\boxed{A^3+A^T}.$

7. Checking options:

• A: $A^2+A^T$ — incorrect
• B: $A^3+I$ — incorrect
• C: $A^3+A^T$ — correct
• D: $A^2+I$ — incorrect

Therefore the correct option is C.