To multiply \(\sqrt{6}\) and \(\sqrt{3b^5}\), you can use the property of square roots that states:
\[ \sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b} \]
So:
\[ \sqrt{6} \cdot \sqrt{3b^5} = \sqrt{6 \cdot 3b^5} \]
Now, let's multiply \(6\) and \(3\):
\[ 6 \cdot 3 = 18 \]
Thus:
\[ \sqrt{6 \cdot 3b^5} = \sqrt{18b^5} \]
Next, we can simplify \(\sqrt{18b^5}\). The number \(18\) can be factored as \(9 \cdot 2\), and the square root of \(9\) is \(3\):
\[ \sqrt{18b^5} = \sqrt{9 \cdot 2 \cdot b^5} \]
Now we can separate the square root:
\[ \sqrt{18b^5} = \sqrt{9} \cdot \sqrt{2} \cdot \sqrt{b^5} \]
This simplifies to:
\[ 3 \cdot \sqrt{2} \cdot \sqrt{b^5} \]
Now, to simplify \(\sqrt{b^5}\), we recognize that:
\[ b^5 = b^4 \cdot b = (b^2)^2 \cdot b \]
So:
\[ \sqrt{b^5} = \sqrt{b^4 \cdot b} = \sqrt{b^4} \cdot \sqrt{b} = b^2 \cdot \sqrt{b} \]
Putting it all together, we have:
\[ 3 \cdot \sqrt{2} \cdot b^2 \cdot \sqrt{b} = 3b^2\sqrt{2b} \]
Thus, the final result in simplest form is:
\[ \boxed{3b^2\sqrt{2b}} \]