Analytic function with zero-derivatives

Question

Suppose $f$ is analytic at $z=0$, and $f(0)=f'(0)=f''(0)=0$. I am to show that there is a function $g$ analytic at $0$ such that $f(z)= z^3g(x)$ in a neighborhood of $0$. I really do not know how to approach this question, hence I seek guidance.

Answer 1

Hint: if $f$ is analytic at $z=0$ and $$f(0)=f''(0)=f''(0)=0,$$ then $$ \begin{align} f(z)=& f(0)+f'(0)z+\frac 12 f''(0)z^2+\frac 16 f'''(0)z^3+\ldots \\ =&\frac 16 f'''(0)z^3+\frac{1}{24}f^{(4)}(0)z^4+\frac{1}{120}f^{(5)}(0)z^5+\ldots \\ =&z^3\times\ldots \end{align} $$