Taylor series for $f(x)= \sqrt[5]{3+2x^3}$ at $a=0$

Question

I have to find Taylor series representation $\sum_{n=0}^\infty a_nx^n$ for the function $f(x)=\sqrt[5]{3+2x^3}$ where $a_n=\frac{f^{n}(0)}{n!}$.

The series itself is easy to calculate $(f(x)=\sqrt[5]{3}+\frac{2x^3}{5\cdot \sqrt[5]{3^4}}-\frac{8x^6}{75\cdot \sqrt[5]{3^4}}\dots)$, but I can't find any ways to represent it as a sum. Thanks in advance for any help.

Answer 1

Since you consider $$f(x)=\sqrt[5]{3+2x^3}$$ rewrite it as $$f(x)=\sqrt[5]3\,\sqrt[5]{1+\frac23 x^3}$$ Now let $t=\frac23 x^3$ and consider that $$\sqrt[5]{1+t}=\sum_{n=0}^\infty\binom{\frac{1}{5}}{n}t^n$$ Back to $x$ and $f(x)$ $$f(x)=\sqrt[5]3\,\sum_{n=0}^\infty\binom{\frac{1}{5}}{n}\left(\frac{2}{3}\right)^n\,x^{3n}$$

Answer 2

Hint: write $$f(x)=\sqrt[5]{3+2x^3}=\sqrt[5]{3}\sqrt[5]{1+\frac{2}{3}x^3}$$ then use the Taylor series for $(1+x)^\alpha$.