Binomial expansions question

Question

In a physics book the autor make the following expansions, given the fact that $z>>d$ (much greater). However I didn't understand how he manage to get the final expression.

$$(z-d/2)^{-3}\approx z^{-3}-3z^{-4}(-d/2)$$ $$(z+d/2)^{-3}\approx z^{-3}-3z^{-4}(-d/2)$$

Edit: In the book it was said that binomial expansion was used.

Answer 1

He factored out $z$ and used the Taylor polynomial of order $1$ for $(1+u)^{-3}$: $$(1+u)^{-3}=1-3u+o(u)$$ Her this gives: $$\Bigl(z-\dfrac d2\Bigl)^{-3}=z^{-3}\Bigl(1-\dfrac d{2z}\Bigl)^{-3}=z^{-3}\Bigl(1+3\dfrac d{2z}\Bigl)+o\Bigl(\frac dz\Bigr)=z^{-3}+3\frac{dz^{-4}}2+o(dz^{-4}).$$

Answer 2

Maclaurin series of $(1 \pm x)^{-3}$ with $x = d/(2z)$.