<s>[INST] Find $f^{(80)}(27)$ where $f(x)=(x+3)^{\frac{1}{3}}\cdot (x-27)$ 
Suppose that $f(x)=(x+3)^{\frac{1}{3}}\cdot (x-27)$.  Use a Taylor series expansion to find $f^{(80)}(27)$.

I tried the following: 
\begin{align}
f'(x) &= (x+3)^{\frac{1}{3}}\cdot 1+(x-27)\cdot \frac{1}{3}(x+3)^{\frac{-2}{3}}\\
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f''(x)
&= \frac{1}{3}(x+3)^{-\frac{2}{3}}+\frac{1}{3}(x-27)\cdot -\frac{2}{3}(x+3)^{-\frac{5}{3}}+\frac{1}{3}(x+3)^{-\frac{2}{3}} \\
&= \frac{2}{3}(x+3)^{-\frac{2}{3}}-\frac{1\cdot 2}{3^2}(x+3)^{-\frac{5}{3}}(x-27)\\
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f'''(x) &=  -\frac{2^2}{3^3}(x+3)^{-\frac{5}{3}}-\frac{1\cdot 2}{3^2}(x+3)^{-\frac{5}{3}}-\frac{1\cdot 2\cdot 5}{3^3}(x+3)^{-\frac{8}{3}}
\end{align}
This is very nasty, please help me to solve in some easy way.
 [/INST] We first expand $f(x)$ at $x=27$. Because it already has a factor $(x-27)$,we just need to expand $(x+3)^{1/3}:=g(x)$. 
Do the Taylor expansion:
$$
g(x) = g(27)+g'(27)(x-27)+\cdots+\frac{1}{79!}g^{(79)}(27)(x-27)^{79}+o((x-27)^{79})
$$
Then the Taylor expansion of $f(x)$ should be (according to the uniqueness of Taylor series):
$$
f(x) = g(27)(x-27)+g'(27)(x-27)^2+\cdots+\frac{1}{79!}g^{(79)}(27)(x-27)^{80}+o((x-27)^{80})
$$
So $f^{(80)}(27)= 80!\times\frac{1}{79!}g^{(79)}(27)=80g^{(79)}(27)$
And 
$$
g^{(79)}(27) = \frac{1}{3}(\frac{1}{3}-1)\cdots(\frac{1}{3}-78)(27+3)^{\frac{1}{3}-79}
$$
Then it's solved. You can calculate it.
</s>