1. Find the product in simplest radical form: $\sqrt{66} \cdot \sqrt{18}$. We use the property $\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$. $$\sqrt{66} \cdot \sqrt{18} = \sqrt{66 \times 18} = \sqrt{1188}$$ Next, factor 1188 to simplify the radical: $$1188 = 4 \times 297 = 4 \times 9 \times 33$$ So, $$\sqrt{1188} = \sqrt{4 \times 9 \times 33} = \sqrt{4} \cdot \sqrt{9} \cdot \sqrt{33} = 2 \cdot 3 \cdot \sqrt{33} = 6\sqrt{33}$$ 2. Find the product in simplest radical form: $\sqrt{6} \cdot \sqrt{21}$. Using the same property: $$\sqrt{6} \cdot \sqrt{21} = \sqrt{6 \times 21} = \sqrt{126}$$ Factor 126: $$126 = 9 \times 14$$ Simplify: $$\sqrt{126} = \sqrt{9 \times 14} = \sqrt{9} \cdot \sqrt{14} = 3\sqrt{14}$$ 3. Find the product in simplest radical form: $\sqrt{3x^{14}} \cdot \sqrt{9x^7}$. Combine under one radical: $$\sqrt{3x^{14}} \cdot \sqrt{9x^7} = \sqrt{3x^{14} \times 9x^7} = \sqrt{27x^{21}}$$ Factor 27: $$27 = 9 \times 3$$ So, $$\sqrt{27x^{21}} = \sqrt{9 \times 3 \times x^{21}} = \sqrt{9} \cdot \sqrt{3} \cdot \sqrt{x^{21}} = 3 \cdot \sqrt{3} \cdot x^{\frac{21}{2}}$$ Recall $\sqrt{x^{21}} = x^{\frac{21}{2}} = x^{10} \cdot x^{\frac{1}{2}} = x^{10} \sqrt{x}$. Therefore, $$3 \cdot \sqrt{3} \cdot x^{10} \sqrt{x} = 3x^{10} \sqrt{3x}$$ Final answers: 1. $6\sqrt{33}$ 2. $3\sqrt{14}$ 3. $3x^{10}\sqrt{3x}$