Closed-forms of infinite series with factorial in the denominator How to evaluate the closed-forms of series

\begin{equation}
1)\,\, \sum_{n=0}^\infty\frac{1}{(3n)!}\qquad\left|\qquad2)\,\, \sum_{n=0}^\infty\frac{1}{(3n+1)!}\qquad\right|\qquad3)\,\, \sum_{n=0}^\infty\frac{1}{(3n+2)!}\\
\end{equation}

Of course Wolfram Alpha can give us the closed-forms
\begin{align}
\sum_{n=0}^\infty\frac{1}{(3n)!}&=\frac{e}{3}+\frac{2\cos\left(\frac{\sqrt{3}}{2}\right)}{3\sqrt{e}}\\
\sum_{n=0}^\infty\frac{1}{(3n+1)!}&=\frac{e}{3}+\frac{2\sin\left(\frac{\sqrt{3}}{2}-\frac{\pi}{6}\right)}{3\sqrt{e}}\\
\sum_{n=0}^\infty\frac{1}{(3n+2)!}&=\frac{e}{3}-\frac{2\sin\left(\frac{\sqrt{3}}{2}+\frac{\pi}{6}\right)}{3\sqrt{e}}
\end{align}
but how to get those closed-forms by hand? I can only notice that
\begin{equation}
\sum_{n=0}^\infty\frac{1}{n!}=\sum_{n=0}^\infty\frac{1}{(3n)!}+\sum_{n=0}^\infty\frac{1}{(3n+1)!}+\sum_{n=0}^\infty\frac{1}{(3n+2)!}=e
\end{equation}
Could anyone here please help me? Any help would be greatly appreciated. Thank you.
PS: Please don't work backward.

Let: $f(0)=1, f'(0)=0, f''(0)=0,f'''(x)=f(x)$
$$f(x) = \sum_{n=0}^{\infty}\frac{x^{3n}}{(3n)!}$$
$$f(x) = \frac{e^x + e^{ux} + e^{vx}} {3} $$ $$u = -1/2+i\sqrt{3}/2; v= u^2 = -1/2-i\sqrt{3}/2$$
$$ f(x) = \frac{e^x}{3}+\frac{e^{-x/2}}{3}\cdot (e^{i\sqrt{3}/2}+e^{-i\sqrt{3}/2})$$
$$ f(x) = \frac{e^x}{3}+\frac{e^{-x/2}}{3}\cdot 2cos((\sqrt{3}/2) x)$$
$$ f(1) = \frac{e}{3}+\frac{2 cos(\frac{\sqrt{3}}{2})}{3 \sqrt{e}}$$
