Let $c=\frac{2\pi}{7}$. What is the value of $\dfrac{\sin3c\cdot\sin6c\cdot\sin9c}{\sin c\cdot\sin2c\cdot\sin3c}$?

1. Since $7c=2\pi$, we can reduce the large angles: $\sin6c=\sin(2\pi-c)=-\sin c$ and $\sin9c=\sin(2\pi+2c)=\sin2c$.
2. The numerator becomes $\sin3c\cdot(-\sin c)\cdot\sin2c=-\sin c\,\sin2c\,\sin3c$.
3. The denominator is $\sin c\,\sin2c\,\sin3c$, which is not zero because $c$, $2c$, $3c$ are $\frac{2\pi}7,\frac{4\pi}7,\frac{6\pi}7$, all strictly between $0$ and $\pi$.
4. Therefore the ratio is $\dfrac{-\sin c\,\sin2c\,\sin3c}{\sin c\,\sin2c\,\sin3c}=\boxed{-1}$.