Finding Taylor series of $x^{-3}$ about $x=a$

Question

How to find the Taylor series of $x^{-3}$ about $x = a$?

Usually I can do ones where $f(x) = (x+c)^{-3}$ but when $c=0$, I'm unsure. Even for positive exponents there's a simple way.

Answer 1

Let $f(x)=x^{-3}$ then $f'(x)=-3x^{-4}$ and $f''(x)=(-3)\times(-4) x^{-5}$ and by induction $$f^{(n)}(x)=(-1)^n\frac{(n+2)!}{2}x^{-(n+3)}$$ so the Taylor series is

$$f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n$$