How does one find the Taylor Series of $ f(x) = \frac{x^{3}}{x+1} $? I know this is relatable to the geometric series:
$$\sum_{n=0}^{\infty} x^{n} = f(x) = \frac{1}{1-x} = 1 + x + x^{2} + x^{3} + x^{4} + \ldots$$
But not sure what should or could be done in order to change the geometric series or the one in question to make the function into a Taylor Series, and an expand it  like the geometric series.
So I know that I can make the function in question look more like a geometric series:
$f(x) = \frac{x^{3}}{1-(-x)}$
but this is still not the right geometric form because of the $x^{3}$ on top.
Just feeling stuck here. Not sure if I should change the original geometric series to look more like the function in question, or to change the function in question to look more like the geometric series.
please help
Thank you
 A: We have
$$\frac {1-x^n}{1-x}=$$
$$1+x+x^2+x^3+x^4....+x^{n-1}$$ and
when $x $  satisfies the condition: $|x|<1,$
we have $\lim_{n\to+\infty}x^n=0$ and
$$\frac{1}{1-x}=\sum_{k=0}^{+\infty}x^k .$$
As $|x|<1\implies |-x|<1$,
WE CAN REPLACE $x $ by $-x $ and get
$$\frac{1}{1-(-x)}=1-x+x^2-x^3+x^4-... $$
and
$$x^3\frac {1}{1+x}=x^3-x^4+x^5-... $$
$$=\sum_{n=3}^{+\infty}(-1)^{n+1}x^n $$
A: An alternative approach. Let $$f(x):=\left(\frac{x^3}{x+1}\right)$$
and using the general Leibniz rule for the product of derivatives we have that
$$\partial^nf(x)=\sum_{k=0}^n\binom{n}{k}\partial^k(x^3)\partial^{n-k}(x+1)^{-1}=\sum_{k=0}^n\binom{n}{k}3^{\underline k}x^{3-k}(-1)^{\overline{n-k}}(x+1)^{-1-n+k}=\\=\frac{(-1)^nx^3 n!}{(x+1)^{n+1}}+\frac{(-1)^{n-1}3x^2 n!}{(x+1)^n}+\frac{(-1)^{n-2}3x n!}{(x+1)^{n-1}}+\frac{(-1)^{n-3}n!}{(x+1)^{n-2}}+0+...$$
Then from above you can search closed expressions for $f^{(k)}(a)$ to define the Taylor expansion of $f$ around $x=a$.
For example for $x=0$ we have that
$$f^{(k)}(0)=\begin{cases}(-1)^{k+1}k!,&k\ge 3\\0,&\text{otherwise}\end{cases}$$
Hence
$$\mathcal T(f,0)=\sum_{k=3}^\infty(-1)^{k+1}x^k,\quad |x|<1$$
