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When $\sqrt[3]{112}$ is simplified, the result is $a \sqrt[3]{b}$, where a is an integer, and b is a positive integer. If b is as small as possible, then what is a + b?

 Apr 4, 2021
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The prime factoriztion of 112 is 2^4*7, so we can factor out 2^3 and get \(\sqrt[3]{112}=2\sqrt[3]{\frac{112}{2^3}}=2\sqrt[3]{14}\). So the answer would be 2 + 14 = 16.

 Apr 4, 2021

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