h) \( \sqrt[5]{160 a^{8}} \cdot \sqrt[4]{81 a^{7} b^{9}} \)
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Beyond the Answer
To simplify \( \sqrt[5]{160 a^{8}} \cdot \sqrt[4]{81 a^{7} b^{9}} \), we'll start by breaking down the components. First, reduce each radical term: 1) For \( \sqrt[5]{160 a^{8}} \): - Break down \( 160 = 32 \times 5 = 2^5 \times 5 \). Thus, \( \sqrt[5]{160} = \sqrt[5]{2^5 \cdot 5} = 2 \sqrt[5]{5} \). - For \( a^{8} \), we have \( \sqrt[5]{a^{8}} = a^{8/5} = a^{1.6} = a^{1}a^{0.6} = a \sqrt[5]{a^3} \). Therefore, \( \sqrt[5]{160 a^{8}} = 2 a \sqrt[5]{5 a^{3}} \). 2) For \( \sqrt[4]{81 a^{7} b^{9}} \): - Break down \( 81 = 3^4 \). Thus, \( \sqrt[4]{81} = 3 \). - For \( a^{7} \), we have \( \sqrt[4]{a^{7}} = a^{7/4} = a^{1.75} = a^{1}a^{0.75} = a \sqrt[4]{a^3} \). - For \( b^{9} \), we have \( \sqrt[4]{b^{9}} = b^{9/4} = b^{2}b^{0.25} = b^{2} \sqrt[4]{b} \). Therefore, \( \sqrt[4]{81 a^{7} b^{9}} = 3 a b^{2} \sqrt[4]{a^3 b} \). Now, substituting back in gives us: \[ \sqrt[5]{160 a^{8}} \cdot \sqrt[4]{81 a^{7} b^{9}} = (2 a \sqrt[5]{5 a^{3}})(3 a b^{2} \sqrt[4]{a^3 b}) \] Next, multiplying the coefficients and like terms: \[ = 6 a^{2} b^{2} \sqrt[5]{5 a^{3}} \sqrt[4]{a^3 b} \] This is a good simplification step! This can be further reduced numerically by multiplying the radicals if required, but the result above captures the essence of the calculation. So, the final answer is: \[ 6a^{2}b^{2} \sqrt[5]{5 a^{3}} \sqrt[4]{a^3 b} \]
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