Let A be a square matrix such that $$AA^T = I$$. Then $$\frac{1}{2}A\left[(A + A^T)^2 + (A - A^T)^2\right]$$ is equal to

$$A$$ is a square matrix with $$AA^T = I$$, i.e., $$A$$ is orthogonal, so $$A^T = A^{-1}$$.

Find $$\frac{1}{2}A\left[(A+A^T)^2 + (A-A^T)^2\right]$$.

Expand the squares.

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

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

Add the two expressions.

$$(A+A^T)^2 + (A-A^T)^2 = 2A^2 + 2(A^T)^2$$

Multiply by $$\frac{1}{2}A$$.

$$\frac{1}{2}A \cdot [2A^2 + 2(A^T)^2] = A \cdot [A^2 + (A^T)^2] = A^3 + A(A^T)^2$$

Simplify $$A(A^T)^2$$.

$$A(A^T)^2 = (AA^T) \cdot A^T = I \cdot A^T = A^T$$

Final result.

$$\frac{1}{2}A[(A+A^T)^2 + (A-A^T)^2] = A^3 + A^T$$

The correct answer is Option 4: $$A^3 + A^T$$.