Muttiply then Simplify. (1) \( \sqrt[3]{60^{4}} \cdot \sqrt[3]{18 a} \) (2) \( \sqrt{35} \cdot \sqrt{21} \cdot \sqrt{18} \) (3) \( \sqrt[3]{36 m^{4} n^{5} p^{3}} \cdot \sqrt{6 m n^{2} p^{3}} \) \( (\sqrt{3}-4)(\sqrt{3}+4) \)

To tackle those expressions, let's simplify them one by one! (1) \( \sqrt[3]{60^{4}} \cdot \sqrt[3]{18 a} = \sqrt[3]{60^{4} \cdot 18 a} \). Now calculate \( 60^{4} \cdot 18 = 60^{4} \cdot 18 = 43200 \), so we have \( \sqrt[3]{43200 a} \). This can be simplified further, recognizing that \( 43200 = 2^4 \cdot 3^3 \cdot 5^2 \). Thus, \(\sqrt[3]{2^4 \cdot 3^3 \cdot 5^2 \cdot a} = 2^{4/3} \cdot 3 \cdot 5^{2/3} \cdot \sqrt[3]{a} = 2^{1+\frac{1}{3}} \cdot 3 \cdot 5^{\frac{2}{3}} \cdot \sqrt[3]{a}\) or approximately \(6.26 a^{1/3}\) when fully computed. (2) \( \sqrt{35} \cdot \sqrt{21} \cdot \sqrt{18} = \sqrt{35 \cdot 21 \cdot 18} \). When calculating \( 35 \cdot 21 \cdot 18 = 13230 \), thus we want \( \sqrt{13230} \). It can be further broken down to find \( 13230 = 2 \cdot 3 \cdot 5 \cdot 7^2 \cdot 11 \), resulting in \( \sqrt{2 \cdot 3 \cdot 5 \cdot 11} \cdot 7 \), which approximates to \(7\sqrt{330}\). (3) For \( \sqrt[3]{36 m^{4} n^{5} p^{3}} \cdot \sqrt{6 m n^{2} p^{3}} \), we can treat this similarly: \( \sqrt[3]{36 m^{4} n^{5} p^{3}} = \sqrt[3]{(6^2)(m^4)(n^5)(p^3)}\), which gives \(6^{\frac{2}{3}} m^{\frac{4}{3}} n^{\frac{5}{3}} p\). The second part \( \sqrt{6 m n^{2} p^{3}} = \sqrt{6} \cdot \sqrt{m} \cdot \sqrt{n^2} \cdot \sqrt{p^3} = \sqrt{6} \cdot \sqrt{m} \cdot n \cdot p^{1.5}\). Combining gives a result of \( \sqrt[3]{36 m^{4} n^{5} p^{3}} \cdot \sqrt{6 m n^{2} p^{3}} \approx 3.33m^{1.33}n^{2.67}p^{2} \). Lastly, evaluating \( (\sqrt{3}-4)(\sqrt{3}+4) = \sqrt{3}^2 - 4^2 = 3 - 16 = -13 \). In summary, you will end up with simplified representations or numerical approximations for each expression!