How to write $(x-1)^3 $ as a Taylor series centred at $x=0$?

Question

I am trying to do a practice question and it is find the Taylor series of $(x-1)^3$ centred at $x=0$ and to show it converges using the definition of radius of convergence on (-R,R) $\forall R \in \mathbb{R}$. The problem that I am having is that if you do it the $$f^{(n)} (0) $$ way then the numbers terminate. So what I was thinking is to do the following: $$ (x-1)^3 = (-1)^3 (1-x)^3 \\ \implies -(1-x)^3 = -\frac{(1-x)^4}{(1-x)} \\ \implies -(1-x)^4 \sum_{n=0}^{\infty} x^{n} $$ but the problem that I am having is that I am not sure if this is the right way to go about doing it because $\sum_{n=0}^{\infty} x^{n}$ is only convergent for $x \in (-1,1) $.

Any help/guidance would be greatly appreciated. Cheers.

Answer 1

You do not even need to do derivatives:

$$(x-1)^3=(x-1)(x^2-2x+1)=x^3-3x^2+3x-1$$

Now we have the expansion centered around $0$.

Answer 2

Just binomial expand:

$$(x-1)^3=x^3-3x^2+3x-1$$

which is the Taylor expansion about $x=0$ with infinite radius of convergence.

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On the side, it is noted that there is no problem with $f^{(n)}(0)=0$.