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If $a$ and $b$ are positive integers and $$(a^\frac{1}{2}\times b^\frac{1}{3})^6=432$$ What is the value of $ab$?

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Note that $$432 = 2^4\cdot 3^3 = 3^3\cdot4^2$$ and the LHS can be written as $$(a^\frac{1}{2} \times b^\frac{1}{3})^6 = a^3\cdot b^2$$ So $a = 3, b = 4$ $\Rightarrow ab = 12$.

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  • $\begingroup$ time difference!! $\endgroup$ Jan 12, 2016 at 15:09
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As a hint, I'd suggest:

  • Write out $(a^{\frac{1}{2}}\times b^{\frac{1}{3}})^3$,
  • Factorize $432$ into prime numbers.
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the relation is equivalent to $a^3.b^2=432$ now prime factorizing we get $432=2^4.3^3=4^2.3^3$ thus $a=3,b=4$ thus $ab=12$

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