Expansion of $(1+3x)^{-5/3}$

Question

In ascending powers of $x$, and including the term $x^3$. Attempted with loads of expansion but need to know a more sophisticated way of expanding. $\text{ }$

Answer 1

$$f(x)=\frac{1}{(3 x+1)^{5/3}}$$ derivatives are $$f'(x)=-\frac{5}{(3 x+1)^{8/3}},\;f''(x)=\frac{40}{(3 x+1)^{11/3}},\;f'''(x)=-\frac{440}{(3 x+1)^{14/3}}$$ so we have $f(0)=1;\;f'(0)=-5;\;f''(0)=40;\;f'''(0)=-440$

Taylor expansion at $x=0$ is $$f(x)=f(0)+f'(0)x+\frac{1}{2!}f''(0)x^2+\frac{1}{3!}f''(0)x^3+O(x^4)$$ that is $$f(x)=1-5x+20x^2-\frac{440}{6}x^3+O(x^4)=\color{red}{1-5x+20x^2-\frac{220}{3}x^3+O(x^4)}$$

Hope this helps