Let $f(x)=\frac{1}{x^3+3x^2+3x+5}$, then what is $f^{(99)}(-1)$? 
Let $f(x)=\frac{1}{x^3+3x^2+3x+5}$, then what is $f^{(99)}(-1)$?

By letting $g(x)=x^3+3x^2+3x+5$, it is easy to see that $g'(-1) = g''(-1) = 0$, and so $f'(-1) = f''(-1) = f'''(-1) = 0$.
However, I have no idea how to compute $f^{(n)}(-1)$ for any positive integer $n$, and in particular for $n=99$.
Any idea would be appreciated.

Note that in a neighbourhood of $x=-1$,
$$f(x)=\frac{1}{x^3+3x^2+3x+5}=\frac{1/4}{1+\frac{(x+1)^3}{4}}=\sum_{n=0}^{\infty}\frac{f^{(n)}(-1)}{n!}(x+1)^n.$$
Now recall that for $|t|<1$,
$$\frac{1}{1+t}=\sum_{n=0}^{\infty}(-1)^n t^n.$$
Can you take it form here?
