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I am trying to solve $27^\frac23$ without a calculator and so far I did this: $$27^\frac23=27^{\frac13+\frac13}=27^\frac13\cdot27^\frac13=\sqrt[3]{27}\cdot\sqrt[3]{27}=3\cdot3=9$$ My book, however, presents a different way of solving this: $$27^\frac23=\sqrt[3]{27^2}=\sqrt[3]{3^6}=3^2=9$$ I know that $27^\frac23=\sqrt[3]{27^2}$ because having an exponent of $\frac23$ is the same as calculating the square of the base and then the cubic root, or the other way around.

What I don't understand is how 27 becomes $3^6$. So, my questions are:

  1. In my attempt I knew that the cube root of 27 is 3, so solving this was easy. But what if there was a more complicated number instead of 27? Would I be able to solve the problem this way?
  2. What was the thought process used to convert 27 into $3^6$?
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    $\begingroup$ Actually $27^2$ becomes $3^6$ $\endgroup$ – arberavdullahu Oct 29 '16 at 15:38
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    $\begingroup$ Expanding a little on the previous comment, $27^2=(3^3)^2=3^{3\cdot 2}=3^6$. $\endgroup$ – Brian M. Scott Oct 29 '16 at 15:39
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    $\begingroup$ @arberavdullahu answered question 2. For question 1, there might not be an obvious solution. This calculation was written with easy numbers to test your understanding of fractional exponents. You did well. $\endgroup$ – Ethan Bolker Oct 29 '16 at 15:41
  • $\begingroup$ Your and the books method are essentially the same albeit with steps in slightly different order. As for thought process of thinking $27^2 = 3^6$... Well, you want to take $\sqrt[3]{27^2}$ so you want to think can we get $27^2$ as something to the power of three, and as $27 = 3^3$ we can get $27^2 = (3^3)^2 = 3^6$. My personal thought process would have gone $27^{2/3} = \sqrt[3]{27^2} = \sqrt[3]{(3^3)^2} = \sqrt[3]{(3^2)^3} = 3^2$. $\endgroup$ – fleablood Oct 29 '16 at 15:59
  • $\begingroup$ You just get good with these in your head with some practice IMO $\endgroup$ – Simply Beautiful Art Oct 31 '16 at 23:36
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There are many ways to think this one through. All will end with the answer $9$. Any textbook can only show one possible answer.

The way the textbook did it was:

$$\sqrt[3]{27^2}=\sqrt[3]{(3^3)^2}=\sqrt[3]{3^6}=3^2=9$$

Alternatively I would have done it as:

$$27^{\frac{2}{3}}=\left(\sqrt[3]{27}\right)^2=\left(\sqrt[3]{3^3}\right)^2=3^2=9$$

As for how to do similar questions I can only recommend knowing or suspecting when a number is a special number. For example if you see $343^{\frac{2}{3}}$ then you are safe to guess that $343$ is a cube of something and work backwards from there.

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    $\begingroup$ Yet another alternative is $27^{\frac23}=(3^3)^{\frac23}=3^2=9$. $\endgroup$ – Ruslan Oct 29 '16 at 15:44
  • $\begingroup$ Actually now that you write that Ruslan, that is how I would have done it. Radical symbols are best avoided. $\endgroup$ – Ian Miller Oct 29 '16 at 15:45
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$27^\frac{2}{3}=(3\cdot\ 3\cdot\ 3)^{\frac{2}{3}}=(3^3)^{\frac{2}{3}}=(3)^{3\cdot\frac{2}{3}}=(3)^{\frac{6}{3}}=3^2$

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Another alternative : (3$^2$)$^{2/3}$.3$^{2/3}$ = 3$^{4/3}$.3$^{2/3}$ = 3$^{6/3}$ = 3$^2$ = $9$

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SIMPLE

$27^\frac{2}{3}=(3\cdot\ 3\cdot\ 3)^{\frac{2}{3}}=(3^3)^{\frac{2}{3}}=(3)^{3\cdot\frac{2}{3}}=**(3)^{\frac{6}{3}}**=3^2$

ACTUALLY. $27^2$. BECOMES $3^6$

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