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}}\\
%
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)\\
%
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.
 A: 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 $n\ge 1$).
A: It is clear that,
$$(x+3)^{\frac{1}{3}} = 30^{\frac{1}{3}}[1+\frac{(x-27)}{30}]^{\frac{1}{3}}.$$
Therefore,
$$(x+3)^{\frac{1}{3}} = 30^{\frac{1}{3}}\sum_{k=0}^{\infty}\frac{\Gamma(\frac{4}{3})}{\Gamma(k+1)\Gamma(\frac{4}{3}-k)}\frac{(x-27)^{k}}{30^{k}},$$
and,
$$f(x) = 30^{\frac{1}{3}}\sum_{k=0}^{\infty}\frac{\Gamma(\frac{4}{3})}{\Gamma(k+1)\Gamma(\frac{4}{3}-k)}\frac{(x-27)^{k+1}}{30^{k}}.$$
Now it is easy to see that:
$f^{80}(27) = 30^{\frac{1}{3}}\frac{\Gamma(\frac{4}{3})}{\Gamma(80)\Gamma(\frac{-233}{3})}\frac{\Gamma (81)}{30^{79}}=30^{\frac{1}{3}}\frac{\Gamma(\frac{4}{3})}{\Gamma(\frac{-233}{3})}\frac{80}{30^{79}}.$
A: 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.
A: 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}
A: Let $g(x)=f(x-3)$. Then $g(x)=x^{\frac13}(x+30)$
The Taylor Series of $x^\frac13$ is
$$\begin{split}x^{\frac13}&=a^{\frac13}+\sum^\infty_{n=1}\frac{a^{\frac13-n}\prod^n_{m=1}\big(\frac43-n\big)}{n!}(x-a)^n\\&=a^\frac13+\frac13a^{-\frac23}(x-a)+...+\frac{a^{-\frac{236}3}\prod^{79}_{n=1}\big(\frac43-n\big)}{79!}(x-a)^{79}+ \frac{a^{-\frac{239}3}\prod^{80}_{n=1}\big(\frac43-n\big)}{80!}(x-a)^{80}+...\end{split}$$
So, we put $a=30$ the degree $80$ term of the Taylor series of $g(x)$ is
$$\begin{split}\frac{(x-30)^{80}}{80!}g^{(80)}(30)&=\deg_{80}\big(x^{\frac13}(x-30+60)\big)\\&
=\frac1{80!}\Bigg(79{\bigg(30^{-\frac{236}3}\bigg)\prod^{79}_{n=1}\bigg(\frac4{3}-n\bigg)}(x-30)^{80}+60\bigg({30^{-\frac{239}3}\bigg)\prod^{80}_{n=1}\bigg(\frac4{3}-n\bigg)}(x-30)^{80}\Bigg)
\\&=\frac1{80!}\Bigg(79{\bigg(30^{-\frac{236}3}\bigg)\prod^{79}_{n=1}\bigg(\frac4{3}-n\bigg)}(x-30)^{80}-\frac{472}3\bigg({30^{-\frac{236}3}\bigg)\prod^{79}_{n=1}\bigg(\frac4{3}-n\bigg)}(x-30)^{80}\Bigg)
\\&=\frac{(x-30)^{80}}{80!}\Bigg(-\frac{235}3\bigg({30^{-\frac{236}3}\bigg)\prod^{79}_{n=1}\bigg(\frac4{3}-n\bigg)}\Bigg)\end{split}$$
So,
$$f^{(80)}(27)=g^{(80)}(30)=-\frac{235}3\bigg({30^{-\frac{236}3}\bigg)\prod^{79}_{n=1}\bigg(\frac4{3}-n\bigg)}$$
