<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] Attention: I suddenly found the OP requires the use of Taylor series expansion. My solution does not meet the requirement. 
We have 
$$f(x) = (x+3)^{1/3}(x-27) = (x+3)^{4/3} - 30(x+3)^{1/3}.$$
Thus, we have, for $k\ge 1$,
\begin{align}
f^{(k)}(x) &= \tfrac{4}{3}(\tfrac{4}{3}-1)(\tfrac{4}{3}-2) \cdots (\tfrac{4}{3}-(k-1))(x+3)^{4/3-k}\\
&\quad - 30\cdot \tfrac{1}{3}(\tfrac{1}{3}-1)(\tfrac{1}{3}-2) \cdots (\tfrac{1}{3}-(k-1))(x+3)^{1/3-k}.
\end{align}
</s>