To find the product of \( \sqrt{6} \) and \( 9\sqrt{12} \), we can start by rewriting the expression:
\[ \sqrt{6} \cdot 9\sqrt{12} \]
First, we can factor and simplify the radical \( \sqrt{12} \):
\[ \sqrt{12} = \sqrt{4 \cdot 3} = \sqrt{4} \cdot \sqrt{3} = 2\sqrt{3} \]
Now we substitute this back into the expression:
\[ \sqrt{6} \cdot 9\sqrt{12} = \sqrt{6} \cdot 9(2\sqrt{3}) = \sqrt{6} \cdot 18\sqrt{3} \]
Next, we can rearrange the terms:
\[ \sqrt{6} \cdot 18\sqrt{3} = 18(\sqrt{6} \cdot \sqrt{3}) = 18\sqrt{6 \cdot 3} \]
Calculating \( 6 \cdot 3 \):
\[ 6 \cdot 3 = 18 \]
So now we have:
\[ 18\sqrt{18} \]
We can simplify \( \sqrt{18} \) further:
\[ \sqrt{18} = \sqrt{9 \cdot 2} = \sqrt{9} \cdot \sqrt{2} = 3\sqrt{2} \]
Thus, substituting back gives us:
\[ 18\sqrt{18} = 18(3\sqrt{2}) = 54\sqrt{2} \]
Therefore, the product of \( \sqrt{6} \) and \( 9\sqrt{12} \) in simplest radical form is:
\[ \boxed{54\sqrt{2}} \]
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