If $1,x_1,x_2,x_3$ are the roots of $x^4-1=0$ and $\omega$ is a complex cube root of unity, the value of $\frac{\left({\omega }^2-x_1\right)\left({\omega }^2-x_2\right)\left({\omega }^2-x_3\right)}{\left(\omega -x_1\right)\left(\omega -x_2\right)\left(\omega -x_3\right)}$ is

$\begin{array}{l}{x}^4-1=0roots1,x_{1,}{x}_{2,}{x}_3\\ x^4-1=\left(x-1\right)\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\\ now\frac{\left({\omega }^2-x_1\right)\left({\omega }^2-x_2\right)\left({\omega }^2-x_3\right)}{\left({\omega }^{ }-x_1\right)\left({\omega }^{ }-x_2\right)\left({\omega }^{ }-x_3\right)}\\ \frac{\left({\omega }^2-1\right)\left({\omega }^2-x_1\right)\left({\omega }^2-x_2\right)\left({\omega }^2-x_3\right)}{\left(\omega +1\right)\left({\omega }^{ }-1\right)\left({\omega }^{ }-x_1\right)\left({\omega }^{ }-x_2\right)\left({\omega }^{ }-x_3\right)}\\ \frac{{\left({\omega }^2\right)}^4-1}{\left(\omega +1\right)\left({\omega }^4-1\right)}=\frac{\left({\omega }^4+1\right)\left({\omega }^4-1\right)}{\left({\omega }^{ }+1\right)\left({\omega }^4-1\right)}\\ \frac{\omega +1}{\omega +1}=1\end{array}$