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

To solve the first expression \((\sqrt{x}-3 \sqrt{3})(\sqrt{x}+3 \sqrt{3})\), we can recognize it as a difference of squares. The formula for a difference of squares is \(a^2 - b^2\), where \(a = \sqrt{x}\) and \(b = 3\sqrt{3}\). \[ (\sqrt{x})^2 - (3\sqrt{3})^2 = x - 27 \] So, \((\sqrt{x}-3 \sqrt{3})(\sqrt{x}+3 \sqrt{3}) = x - 27\). Now, for the second expression \((\sqrt{x}+\sqrt{3})^{2}\), we can expand it using the formula \((a+b)^2 = a^2 + 2ab + b^2\): \[ (\sqrt{x})^2 + 2(\sqrt{x})(\sqrt{3}) + (\sqrt{3})^2 = x + 2\sqrt{3x} + 3 \] Thus, \((\sqrt{x}+\sqrt{3})^{2} = x + 2\sqrt{3x} + 3\). To summarize: 1. \( (\sqrt{x}-3 \sqrt{3})(\sqrt{x}+3 \sqrt{3}) = x - 27 \) 2. \( (\sqrt{x}+\sqrt{3})^{2} = x + 2\sqrt{3x} + 3 \)