<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] Put $g(x)=(x+3)^{1/3}$. We have $f(x)=g(x)(x-27)$. It suggests that if a Taylor series of $f$ at $x_0=27$ is $\sum_{n=0}^\infty \frac {f^{(n)}(x_0)}{n!}(x-x_0)^n$ and a Taylor series of $g$ at $x_0=27$ is $\sum_{n=0}^\infty \frac {g^{(n)}(x_0)}{n!}(x-x_0)^n$ then  for each $n\ge 0$ we have $$\frac {g^{(n)}(x_0)}{n!}= \frac {f^{(n+1)}(x_0)}{(n+1)!}.$$ 
But we don’t need the Taylor series to show the latter equality. It can be shown directly, by a simple induction or by Leibniz differentiation rule stating that 
$$f^{(n+1)}(x_0)=\sum_{i=0}^{n+1}{n+1\choose i}(x-x_0)^{(i)}g^{(i)}(x)\large|_{x=x_0}=(n+1)g^{(n)}(x_0)=$$ $$(x_0+3)^{\frac 13-n}(n+1)\prod_{i=0}^{n-1} \left(\frac 13-i\right)$$ 
(the last equality holds for any natural $n\ge 1$).
</s>