(3√3-√11)*(√27+√11)

To simplify the expression \((3\sqrt{3} - \sqrt{11})(\sqrt{27} + \sqrt{11})\), we can use the distributive property (also known as the FOIL method for binomials):

1. Expand the expression: \[ (3\sqrt{3})(\sqrt{27}) + (3\sqrt{3})(\sqrt{11}) - (\sqrt{11})(\sqrt{27}) - (\sqrt{11})(\sqrt{11}) \]

2. Calculate each term:

   • For \(3\sqrt{3} \cdot \sqrt{27}\): \[ \sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3} \quad \Rightarrow \quad 3\sqrt{3} \cdot 3\sqrt{3} = 9 \cdot 3 = 27 \]

   • For \(3\sqrt{3} \cdot \sqrt{11}\): \[ 3\sqrt{3}\sqrt{11} = 3\sqrt{33} \]

   • For \(-\sqrt{11} \cdot \sqrt{27}\): \[ -\sqrt{11}\sqrt{27} = -\sqrt{11 \cdot 27} = -\sqrt{297} \]

   • For \(-\sqrt{11} \cdot \sqrt{11}\): \[ -\sqrt{11}\sqrt{11} = -11 \]

3. Put it all together: \[ 27 + 3\sqrt{33} - \sqrt{297} - 11 \]

4. Combine like terms: \[ 27 - 11 = 16 \]

Thus, the expression simplifies to: \[ 16 + 3\sqrt{33} - \sqrt{297} \]

If you need further simplification on \(\sqrt{297}\): \[ \sqrt{297} = \sqrt{3 \cdot 99} = \sqrt{3 \cdot 9 \cdot 11} = 3\sqrt{11} \]

So, replacing \(\sqrt{297}\) gives: \[ 16 + 3\sqrt{33} - 3\sqrt{11} \]

This is the simplified form of your expression: \[ \boxed{16 + 3\sqrt{33} - 3\sqrt{11}} \]