A possible solution without use a Taylor series.Observe that:
$\\ \\ \displaystyle \sin(3\gamma)=\sin(2\gamma)\cos(\gamma)+\sin(\gamma)\cos(2\gamma)=2\sin(\gamma)\cos^2(\gamma)+\sin(\gamma)(1-2\sin^2(\gamma))=2\sin(\gamma)(1-\sin^2\gamma)+\sin(\gamma)(1-2\sin^2(\gamma))=3\sin(\gamma)-4\sin^3(\gamma)\Rightarrow \sin^3(\gamma)=\frac{1}{4}\left(3\sin(\gamma)-\sin(3\gamma)\right)$
\begin{equation}
\sin^3(\gamma)=\frac{1}{4}\left(3\sin(\gamma)-\sin(3\gamma)\right)
\end{equation}
Do it $\displaystyle \gamma=\frac{\phi}{3^k}$:
\begin{equation}
\sin^3\left(\frac{\phi}{3^k}\right)=\frac{1}{4}\left(3\sin\left(\frac{\phi}{3^k}\right)-\sin\left(\frac{\phi}{3^{k-1}}\right)\right)
\end{equation}
Multiplying by $\displaystyle 3^{k-1}$:
\begin{equation}
3^{k-1}\sin^3\left(\frac{\phi}{3^k}\right)=\frac{1}{4}\left(3^{k}\sin\left(\frac{\phi}{3^k}\right)-3^{k-1}\sin\left(\frac{\phi}{3^{k-1}}\right)\right)
\end{equation}
Applying summation on both sides of equality, we will have:
$\\ \\ \displaystyle \sum_{k=1}^{n}3^{k-1}\sin^3\left(\frac{\phi}{3^k}\right)=\sum_{k=1}^{n}\frac{1}{4}\left(3^{k}\sin\left(\frac{\phi}{3^k}\right)-3^{k-1}\sin\left(\frac{\phi}{3^{k-1}}\right)\right)= \frac{1}{4}\left(\left(3\sin\left(\frac{\phi}{3^k}\right)-\sin\left(\phi\right)\right)+\left(3^{2}\sin\left(\frac{\phi}{3^k}\right)-3\sin\left(\frac{\phi}{3}\right)\right)+...+\left(3^{n}\sin\left(\frac{\phi}{3^n}\right)-3^{n-1}\sin\left(\frac{\phi}{3^{n-1}}\right)\right)\right) =\frac{1}{4}\left(3^{n}\sin\left(\frac{\phi}{3^n}\right)-\sin(\phi)\right)\Rightarrow \sum_{k=1}^{n}3^{k-1}\sin^3\left(\frac{\phi}{3^k}\right)=\frac{1}{4}\left(3^{n}\sin\left(\frac{\phi}{3^n}\right)-\sin(\phi)\right)\\ \\$
Take the limit:
\begin{equation*}
\lim_{n\rightarrow \infty}\sum_{k=1}^{n}3^{k-1}\sin^{3}\left(\frac{\phi}{3^{k}}\right)=\frac{1}{4}\left(\phi-\sin(\phi)\right)
\end{equation*}
Notice, on the other hand, using the inequality $ \displaystyle \sin x \leq x $ and using the infinite arithmetic progression formula, follows:
$\\ \displaystyle \sin\left(\frac{\phi}{3^{k}}\right) \leq \frac{\phi}{3^{k}}\Rightarrow \sin^{3}\left(\frac{\phi}{3^{k}}\right) \leq \frac{\phi^3}{3^{3k}} \Rightarrow 3^{k-1}\sin^{3}\left(\frac{\phi}{3^{k}}\right) \leq \frac{\phi^3}{3^{2k+1}}\Rightarrow \sum_{k=1}^{\infty}3^{k-1}\sin^{3}\left(\frac{\phi}{3^{k}}\right) \leq \sum_{k=1}^{\infty} \frac{\phi^3}{3^{2k+1}}=\frac{\phi^3}{3}\sum_{k=1}^{\infty} \frac{1}{3^{2k}}=\frac{\phi^3}{3\times 8}\Rightarrow \sum_{k=1}^{\infty}3^{k-1}\sin^{3}\left(\frac{\phi}{3^{k}}\right) \leq \frac{\phi^3}{3\times 8} \Rightarrow \frac{1}{4}\left(\phi-\sin(\phi)\right) \leq \frac{\phi^3}{3\times 8} \Rightarrow \phi-\frac{\phi^3}{6}\leq \sin(\phi) $