Number Base Systems Equation

$(xxx)_b = x \cdot b^2 + x \cdot b + x = x(b^2 + b + 1)$. $x(b^2 + b + 1) = x^3 \implies b^2 + b + 1 = x^2$. Since $x < b$, $x^2 < b^2 < b^2 + b + 1$, so $b^2 + b + 1 = x^2$ has no valid integer solution with $x < b$. Thus, None of the above.