Simplifying $27^\frac23$ without a calculator 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:


*

*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?

*What was the thought process used to convert 27 into $3^6$?


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