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?

Let $1, w, w^2$ be the cube roots of unity. Let $f(x) = (e^x + e^{xw} + e^{xw^2})/3$. Expand $f$ in a Taylor series. Because the cube roots of unity sum to 0, all terms vanish except where the exponents are multiples of 3, in which case they give the coefficients of your series. You can convert f(x) to $\frac{e^x}{3} +\frac{2e^{-x/2}}{3} cos(\frac{\sqrt(3) x}{2})$ by using $cos(z) = \frac{(e^{iz}+e^{-iz})}{2}$.
