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.

Using that $$a^{-n}=\frac{1}{a^n}$$ we get
$$\frac{x^3}{16\cdot 2y^3}=…$$ and it must be $$x\neq 0$$
