Combine the following expressions. \( \sqrt{48 n^{3}}+\sqrt{9 n}-\sqrt{3 n} \) \( (4 n-1) \sqrt{3 n}+3 \sqrt{n} \) \( 3 n \sqrt{3 n}+3 \sqrt{n} \) \( (4 n-1) \sqrt{3 n}-3 \sqrt{n} \)

Let's combine the given expressions step by step: 1. **Combining the first expression:** \(\sqrt{48 n^{3}}+\sqrt{9 n}-\sqrt{3 n}\) We can simplify \(\sqrt{48 n^3}\) as follows: \(\sqrt{48 n^3} = \sqrt{16 \cdot 3 \cdot n^2 \cdot n} = 4n \sqrt{3n}\). So, we rewrite the expression: \(4n \sqrt{3n} + 3\sqrt{n} - \sqrt{3n}\). This simplifies to: \(4n \sqrt{3n} + (3 - 1)\sqrt{3n} = 4n \sqrt{3n} + 2\sqrt{3n}\). We can factor out \(\sqrt{3n}\): \((4n + 2)\sqrt{3n} = 2(2n + 1)\sqrt{3n}\). 2. **Combining the second expression:** \((4 n-1) \sqrt{3 n}+3 \sqrt{n}\). This remains as is since it is already in its simplest form. 3. **Combining the third expression:** \(3 n \sqrt{3 n} + 3 \sqrt{n}\). This, too, remains as is. 4. **Combining the fourth expression:** \((4 n-1) \sqrt{3 n}-3 \sqrt{n}\). We can expand this as: \(4n \sqrt{3n} - \sqrt{3n} - 3\sqrt{n}\). This simplifies to: \((4n - 1)\sqrt{3n} - 3\sqrt{n}\). Now combining everything: Result = \((2(2n + 1)\sqrt{3n}) + (4 n-1) \sqrt{3 n}+3 \sqrt{n} + (3 n \sqrt{3 n}+3 \sqrt{n}) + (4 n-1) \sqrt{3 n}-3 \sqrt{n}\). Thus, grouping similar terms we have: 1. Combine the \(\sqrt{3n}\) terms: - From the first expression: \(2(2n + 1)\sqrt{3n}\). - From the second expression: \((4n - 1)\sqrt{3n}\). - From the third expression: \(3n\sqrt{3n}\). - From the fourth expression: \((4n - 1)\sqrt{3n}\). This will yield a total of: - \((8n + 2 - 1 + 3n + 4n - 1)\sqrt{3n}\). 2. Combine the \(\sqrt{n}\) terms: - From the second expression: \(3 \sqrt{n}\). - From the third expression: \(3 \sqrt{n}\). - From the fourth expression: \(-3 \sqrt{n}\). This will yield a total of: - \(3 + 3 - 3 = 3\sqrt{n}\). Putting it all together, we get: \((8n + 7)\sqrt{3n} + 3\sqrt{n}\). So the final combined expression is: \[(8n + 7)\sqrt{3n} + 3\sqrt{n}.\]