Finding sum of infinite series $1+\frac{x^3}{3!}+\frac{x^6}{6!}+\frac{x^9}{9!}+\ldots $ So the question is 'express the power series $$1+\frac{x^3}{3!}+\frac{x^6}{6!}+\frac{x^9}{9!}+\ldots $$
 in closed form'.
Now we are allowed to assume the power series of $e^x$ and also we derived the power series for $\cosh x$ using exponentials. 
Now my question is the best way to approach the problem above. There is a hint that says 'use the fact that $\zeta^2+\zeta +1=0 $ where $\zeta $ is a cube root of unity'. 
The way I solved this was to recognise that the third derivative of the series was equal to the series itself. So I just solved a linear ODE of order 3 and then found the constants (and so the hint made sense - in a way). But I don't think the question was designed for me to do this and so I feel as though I'm missing something obvious that makes this problem very easy.
Can anyone see any alternatives that make use of the hint in a more natural way?

We evaluate the power series $e^{\zeta x}$, noting that $\zeta^2 = -(1+\zeta)$ and $\zeta^3 = 1$, and get
$$
e^{\zeta x} = 1 + \zeta x + \frac{\zeta^2x^2}{2!} + \frac{\zeta^3x^3}{3!} + \frac{\zeta^4 x^4}{4!} + \cdots\\
= 1 + \zeta x - \frac{(1+\zeta)x^2}{2!} + \frac{x^3}{3!} + \frac{\zeta x^4}{4!} -\cdots
$$
while evaluating the power series $e^{\zeta^2 x}$ gives
$$
e^{\zeta^2 x} = 1 + \zeta^2 x + \frac{\zeta^4x^2}{2!} + \frac{\zeta^6x^3}{3!} + \frac{\zeta^8 x^4}{4!} + \cdots\\
= 1 - (1+\zeta)x + \frac{\zeta x^2}{2!} + \frac{x^3}{3!} - \frac{(1+\zeta)x^4}{4!} + \cdots
$$
Adding these together gives us
$$
e^{\zeta x} + e^{\zeta^2 x} = 2 - x - \frac{x^2}{2!} + \frac{2x^3}{3!} - \frac{x^4}{4!} - \cdots
$$
Finally, we see that if we add the power series of $e^x$ to this one, we get $3$ times the series we were after. That means that the final answer is
$$
1 + \frac{x^3}{3!} + \frac{x^6}{6!} + \cdots = \frac{e^x + e^{\zeta x} + e^{\zeta^2 x}}{3}
$$
