Multiply and simplify. \( \begin{array}{l}(\sqrt{x}-3 \sqrt{3})(\sqrt{x}+3 \sqrt{3})=\square \\ (\sqrt{x}+\sqrt{3})^{2}=\square\end{array} \)

Let's simplify each expression step by step. 1. **Multiplying \((\sqrt{x} - 3\sqrt{3})(\sqrt{x} + 3\sqrt{3})\):** This is a difference of squares, which follows the formula: \[ (a - b)(a + b) = a^2 - b^2 \] Here, \(a = \sqrt{x}\) and \(b = 3\sqrt{3}\). Applying the formula: \[ (\sqrt{x})^2 - (3\sqrt{3})^2 = x - 9 \times 3 = x - 27 \] So, \[ (\sqrt{x} - 3\sqrt{3})(\sqrt{x} + 3\sqrt{3}) = x - 27 \] 2. **Expanding \((\sqrt{x} + \sqrt{3})^2\):** Use the formula for squaring a binomial: \[ (a + b)^2 = a^2 + 2ab + b^2 \] Here, \(a = \sqrt{x}\) and \(b = \sqrt{3}\). Applying the formula: \[ (\sqrt{x})^2 + 2(\sqrt{x})(\sqrt{3}) + (\sqrt{3})^2 = x + 2\sqrt{3x} + 3 \] So, \[ (\sqrt{x} + \sqrt{3})^2 = x + 2\sqrt{3x} + 3 \] **Final Simplified Results:** \[ \begin{align*} (\sqrt{x} - 3\sqrt{3})(\sqrt{x} + 3\sqrt{3}) &= x - 27 \\ (\sqrt{x} + \sqrt{3})^{2} &= x + 2\sqrt{3x} + 3 \end{align*} \] **Answer:** \[ \begin{align*} (\sqrt{x} - 3\sqrt{3})(\sqrt{x} + 3\sqrt{3}) &= x - 27 \\ (\sqrt{x} + \sqrt{3})^{2} &= x + 2\sqrt{3x} + 3 \end{align*} \]