Multiplication/Division: How did the author simplify $(1/64)z^-3/1-4z$ to the result in the image?

Question

I have tried multiplying the numerator and denominator by 4/z and also dividing both by 4z but alas I can't seem to get the answer. Thank you very much to anyone who answers.

Answer 1

I think it's indeed a little bit wrong. Let's see... $$ \begin{split} \frac{ \frac{1}{64}z^{-3} }{ 1 - 4z } =& \frac{ \frac{z}{4\cdot 16}z^{-4} }{ \frac{z/4}{z/4} - 16\cdot \frac{z}{4} } \\ =&\frac{ \left( \frac{z}{4} \right) \frac{1}{16}z^{-4} }{ \left( \frac{z}{4} \right) \left[ \frac{1}{z/4} - 16 \right] } \\ =&\frac{ \frac{1}{16}z^{-4} }{ \left[ \frac{1}{z/4} - 16 \right] } \\ =&\frac{ \frac{1}{16}z^{-4} }{ 16 \left[ \frac{1}{4}z^{-1} - 1 \right] } \\ \end{split} $$ when $z \neq 0$.