Simplify $\frac{4^{-2}x^3y^{-3}}{2x^0}$ to $\frac{x^3}{32y^3}$ I am to simplify $\frac{4^{-2}x^3y^{-3}}{2x^0}$ and I know that the solution is $\frac{x^3}{32y^3}$
I understand how to apply rules of exponents to individual components of this expression but not as a whole.
For example, I know that $4^{-2}$ = $\frac{1}{4^2}$ = $1/16$
But how can I integrate this 1/16 to the expression?
Do I remove the original $4^{-2}$ and replace with 1 to the numerator and a 16 to the denominator like this?
$\frac{1x^3y^{-3}}{16*2x^0}$
How can I simplify the above expression to $\frac{x^3}{32y^3}$? Would be grateful for a granular set of in between steps, even if they are most basic to others.

Do it step by step :
First, as you said $4^{-2} = \frac{1}{16}$, replace it in the given expression :
$$ \dfrac{4^{-2}x^3y^{-3}}{2x^0} = \dfrac{x^3y^{-3}}{16\cdot 2 x^0}$$
Then, simplify the denominator, $16\cdot 2 = 32$ and $x^0 = 1$ so that : 
$$\dfrac{4^{-2}x^3y^{-3}}{2x^0}  = \dfrac{x^3\cdot y^{-3}}{32} $$
Then, because $y^{-3} = \dfrac{1}{y^3}$, you can find the final expression :
$$ \dfrac{4^{-2}x^3y^{-3}}{2x^0}  = \dfrac{x^3}{32y^3}$$
