5) \( \sqrt{15 x^{3}} \cdot \sqrt{21 x^{3}} \)
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To simplify \( \sqrt{15 x^{3}} \cdot \sqrt{21 x^{3}} \), you can combine the square roots: \[ \sqrt{15 x^{3}} \cdot \sqrt{21 x^{3}} = \sqrt{(15 \cdot 21) \cdot (x^{3} \cdot x^{3})} = \sqrt{315} \cdot \sqrt{x^{6}} = \sqrt{315} \cdot x^{3} \] Now, let's simplify \( \sqrt{315} \). The prime factorization of 315 is \( 3^2 \cdot 5 \cdot 7 \). Therefore: \[ \sqrt{315} = \sqrt{3^2 \cdot 5 \cdot 7} = 3\sqrt{35} \] Putting it all together, you get: \[ \sqrt{15 x^{3}} \cdot \sqrt{21 x^{3}} = 3 \sqrt{35} \cdot x^{3} = 3x^{3} \sqrt{35} \] So, the final answer is: \[ 3x^{3} \sqrt{35} \]