Radical Sum Simplification
Simplify $ \frac{3}{\sqrt[5]{16}}+\frac{1}{\sqrt{3}} $ and rationalize the denominator. The result can be expressed in the form $ \frac{a^2\sqrt[5]{b}+b\sqrt{a}}{ab} $, where $ a $ and $ b $ are integers. What is the value of the sum $ a+b $?
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- $\frac{a}{b}$
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- $a^n$
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Solution
Rationalizing each of the two fractions on its own will make creating a common denominator easier. For the first fraction, if we recognize the denominator $ \sqrt[5]{16} $ as $ \sqrt[5]{2^4} $, then that means multiplying the numerator and denominator by $ \sqrt[5]{2} $ will leave us with 2 in the denominator: $$\frac{3}{\sqrt[5]{16}}\cdot\frac{\sqrt[5]{2}}{\sqrt[5]{2}}=\frac{3\sqrt[5]{2}}{\sqrt[5]{2^5}}=\frac{3\sqrt[5]{2}}{2}.$$For the second fraction, we have $ \frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3} $. Now we find a common denominator: $$\frac{3\sqrt[5]{2}}{2}+\frac{\sqrt{3}}{3}=\frac{9\sqrt[5]{2}+2\sqrt{3}}{6}.$$So, matching our answer with the form in the problem, we get that $ a=3 $ and $ b=2 $, which means $ a+b=\boxed{5} $.
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