To simplify the expression \( \sqrt{63} \cdot \sqrt{18} \), we can use the property of square roots that states:
\[ \sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b} \]
Applying this property, we have:
\[ \sqrt{63} \cdot \sqrt{18} = \sqrt{63 \cdot 18} \]
Next, we need to compute \( 63 \cdot 18 \):
\[ 63 \cdot 18 = 1134 \]
Now, we simplify \( \sqrt{1134} \). We start by factoring \( 1134 \) into its prime factors.
First, divide by \( 2 \):
\[ 1134 \div 2 = 567 \]
Next, continue factoring \( 567 \). Divide by \( 3 \):
\[ 567 \div 3 = 189 \]
Next, continue factoring \( 189 \). Divide by \( 3 \) again:
\[ 189 \div 3 = 63 \]
Now factor \( 63 \):
\[ 63 \div 3 = 21 \]
And factor \( 21 \):
\[ 21 \div 3 = 7 \]
So, we have completely factored \( 1134 \):
\[ 1134 = 2 \cdot 3^3 \cdot 7 \]
Now we apply this factorization to simplify \( \sqrt{1134} \):
\[ \sqrt{1134} = \sqrt{2 \cdot 3^3 \cdot 7} \]
Using the property of square roots, we can separate the perfect squares inside the square root:
\[ \sqrt{1134} = \sqrt{2} \cdot \sqrt{3^2} \cdot \sqrt{3} \cdot \sqrt{7} = \sqrt{2} \cdot 3 \cdot \sqrt{21} \]
Thus, we can write the expression as:
\[ \sqrt{1134} = 3\sqrt{42} \]
Therefore, the final result is:
\[ \sqrt{63} \cdot \sqrt{18} = 3\sqrt{42} \]
Thus, the product in simplest radical form is:
\[ \boxed{3\sqrt{42}} \]
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