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For my precalc class we had the problem $\sqrt 3\sqrt[3]6$ and we had to convert it to rational exponents. I thought it was easy, $3^\frac 12\times6^\frac 13$ but the answer was $2^\frac 13\times3^\frac 56$. Any explanation please?

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    $\begingroup$ $6^{1/3} \ = \ (2 \ \cdot 3)^{1/3} \ = \ 2^{1/3} \ \cdot \ 3^{1/3} \ $ , so $ \ 3^{1/2} \ $ times that is...? $\endgroup$ – colormegone Sep 5 '15 at 21:27
  • $\begingroup$ But I think your answer is also correct, unless there were other specs in the question. $\endgroup$ – Stephen Montgomery-Smith Sep 5 '15 at 21:29
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    $\begingroup$ You will find as you move toward progressively more advanced mathematics that there is often more than one way to express the same number or function, sometimes there are many equivalent ways. When you see an answer given, check to see if your own answer could be equivalent: students sometimes think they've done a problem wrong just because the book's or instructor's answer is different(-looking). $\endgroup$ – colormegone Sep 5 '15 at 21:31
  • $\begingroup$ Oh okay! Thanks! $\endgroup$ – Hannah Everlong Sep 5 '15 at 21:33
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Your answer is correct, but it could be simplified further: $$3^\frac{1}{2}\times6^\frac{1}{3}=3^\frac{1}{2}\times2^\frac{1}{3}\times3^\frac{1}{3}=2^\frac{1}{3}\times3^\frac{5}{6}$$

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As answered before, many equivalent solutions can be found. So you should try to put yourself in your teacher's shoes: gear toward "uniqueness" and simplicy. The $2^{\frac{1}{3}}3^{\frac{5}{6}}$ is somehow simpler because the factors are primes, and are written in increasing order. So you gain in simplicy (in a teacher's point of view) by careful factoring. Your initial guess, though correct, is a mere translation of the root expression, and brings no "added value" to your reasoning.

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