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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.

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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 :

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

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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 $\frac{x^3}{32y^3}$.

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

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