The expression you've provided is:

\[
33 \sqrt{12} \cdot \sqrt{6}
\]

Let's simplify this step by step.

1. **Simplify the square roots**:
   - $\sqrt{12}$ can be simplified as $\sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}$.
   - $\sqrt{6}$ cannot be simplified further.

   So, the expression becomes:
   \[
   33 \times 2\sqrt{3} \times \sqrt{6}
   \]

2. **Combine the square roots**:
   - $\sqrt{3} \times \sqrt{6} = \sqrt{3 \times 6} = \sqrt{18}$.
   - $\sqrt{18}$ can be simplified as $\sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2}$.

   Now, the expression is:
   \[
   33 \times 2 \times 3\sqrt{2} = 33 \times 6\sqrt{2} 
   \]

3. **Multiply the coefficients**:
   \[
   33 \times 6 = 198
   \]

   Thus, the expression simplifies to:
   \[
   198\sqrt{2}
   \]

So, the simplified result is $198\sqrt{2}$.

Would you like any further details or explanations?

Here are 5 related questions:
1. How do you simplify expressions involving square roots?
2. What is the process for multiplying and simplifying square roots?
3. Can you demonstrate how to rationalize a denominator with square roots?
4. How do you handle square roots in fractions?
5. How do you expand binomials involving square roots?

**Tip**: When simplifying square roots, always look for perfect squares within the number to break it down into simpler terms.

Learn how to simplify the expression $$ 33\sqrt{12}\cdot\sqrt{6} $$ step-by-step with detailed explanations. This problem involves simplifying square roots and multiplying radicals, suitable for high school students (Grades 9-12).

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