If $1,{\mathrm{z}}_1,{\mathrm{z}}_2,{\mathrm{z}}_3,\dots ,{\mathrm{z}}_{\mathrm{n}-1}$ be the n, nth roots of unity and $\mathrm{\omega }$ be a non-real complex cube root of unity, then $\prod _{\mathrm{r}=1}^{\mathrm{n}-1} \left(\mathrm{\omega }-{\mathrm{z}}_{\mathrm{r}}\right)$ can be equal to 

Since, $1,{\mathrm{z}}_1,{\mathrm{z}}_2,{\mathrm{z}}_{3.....}{\mathrm{z}}_{\mathrm{n}-1}$ are the n, nth roots of unity.
Therefore, 
$$\begin{gathered} {\mathrm{z}}^{\mathrm{n}}-1=(\mathrm{z}-1)\left(\mathrm{z}-{\mathrm{z}}_1\right)\left(\mathrm{z}-{\mathrm{z}}_2\right)\dots \left(\mathrm{z}-{\mathrm{z}}_{\mathrm{n}-1}\right) \\ \Rightarrow \frac{{\mathrm{z}}^{\mathrm{n}}-1}{\mathrm{z}-1}=\left(\mathrm{z}-{\mathrm{z}}_1\right)\left(\mathrm{z}-{\mathrm{z}}_2\right)\dots \left(\mathrm{z}-{\mathrm{z}}_{\mathrm{n}-1}\right) \\ =\prod _{\mathrm{r}=1}^{\mathrm{n}-1} \left(\mathrm{z}-{\mathrm{z}}_{\mathrm{r}}\right) \end{gathered}$$
Now, putting $\mathrm{z}=\mathrm{\omega }$, we get 
$$\begin{gathered} \prod _{\mathrm{r}=1}^{\mathrm{n}-1} \left(\mathrm{\omega }-{\mathrm{z}}_{\mathrm{r}}\right)=\frac{{\mathrm{\omega }}^{\mathrm{n}}-1}{\mathrm{\omega }-1} \\ =\left\{\begin{array}{cl}0,& \text{ if }\mathrm{n}=3\mathrm{r},\mathrm{r}\in \mathrm{Z}\\ 1,& \text{ if }\mathrm{n}=3\mathrm{r}+1,\mathrm{r}\in \mathrm{Z}\\ 1+\mathrm{\omega },& \text{ if }\mathrm{n}=3\mathrm{r}+2,\mathrm{r}\in \mathrm{Z}\end{array}\right. \end{gathered}$$