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 :

$$\dfrac{4^{-2}x^3y^{-3}}{2x^0} = \dfrac{x^3}{32y^3}$$

You are right in your steps to arrive at $$\frac{1x^3y^{-3}}{16×2x^0}$$. Now $$x^0=1$$, so it can be erased alongside the 1 in the numerator to get $$\frac{x^3y^{-3}}{16×2}$$. Multiplying top and bottom by $$y^3$$, and reducing $$16×2=32$$, yields the desired $$\frac{x^3}{32y^3}$$.

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