Question:

**Step 1: Extract conditions from $A^T A = I$.** The condition $A^T A = I$ implies that the columns of $A$ form an orthonormal set. Let the columns of $A$ be $C_1 = \begin{bmatrix} a \\ b \\ c \end{bmatrix}$, $C_2 = \begin{bmatrix} b \\ c \\ a \end{bmatrix}$, and $C_3 = \begin{bmatrix} c \\ a \\ b \end{bmatrix}$. The dot product of each column with itself must be 1: $C_1 \cdot C_1 = a^2 + b^2 + c^2 = 1$ $C_2 \cdot C_2 = b^2 + c^2 + a^2 = 1$ $C_3 \cdot C_3 = c^2 + a^2 + b^2 = 1$ These equations all yield the same condition: $$a^2 + b^2 + c^2 = 1 \quad (1)$$ The dot product of distinct columns must be 0: $C_1 \cdot C_2 = ab + bc + ca = 0$ $C_1 \cdot C_3 = ac + ba + cb = 0$ $C_2 \cdot C_3 = bc + ca + ab = 0$ These equations all yield the same condition: $$ab + bc + ca = 0 \quad (2)$$ **Step 2: Determine the sum $a+b+c$.** We use the algebraic identity for the square of the sum of three variables: $$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$$ Substitute the values from equations (1) and (2): $$(a+b+c)^2 = 1 + 2(0) = 1$$ Taking the square root of both sides, we get $a+b+c = \pm 1$. Given that $a, b, c$ are positive real numbers, their sum $a+b+c$ must be positive. Therefore, $$a+b+c = 1 \quad (3)$$ **Step 3: Calculate $a^3 + b^3 + c^3$.** We use the algebraic identity for the sum of cubes: $$a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - (ab+bc+ca))$$ Substitute the given condition $abc=1$ and the derived values from equations (1), (2), and (3): $$a^3 + b^3 + c^3 - 3(1) = (1)(1 - (0))$$ $$a^3 + b^3 + c^3 - 3 = 1$$ $$a^3 + b^3 + c^3 = 4$$