In the following factoring question

$\displaystyle 64x^3 - 27y^6$

(a - b) (a^2 + ab + b^2) is used

However, I am stuck right off the bat. The cube root of $\displaystyle 64x^3$ is 4x, but does one do the cubed root of $\displaystyle 27y^6$. Its obviously a difference of perfect cubes because x^3 and y^6 are both multiples of 3. The book tells me the cubed root of $\displaystyle 27^6$ is $\displaystyle 3y^2$, but I do not know specifically how the $\displaystyle y^6 $turns into $\displaystyle y^2$ (obviously the cube root of 27 is 3).

Could some one please clarify and show me?