Question

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

Answer 1

$x^4-1=0$ $\frac{(w^2-x_4)(w^2-x_2)(w^2-x_3)}{(w-x_1)(w-x_2)(w-x_3)}$

$x^2=1,x^2=-1$

$x=1,-1,i,-i$ $=\frac{(w^2-1)(w^2-i)(w^2+i)}{w+1\left)\right(w-i\left)\right(w+i)}$

$1,x_1,x_2,x_3$

$x_1=-1,x_2=i,$ $=\frac{(w^2+1)(w^4+1)}{(w+1)(w^2+1)}$

$x_3=-1$ $=\frac{w^4+1}{w+1}$

$=\frac{w^4+1}{w+1}=\frac{w^3+w+1}{w+1}$

$=\frac{w+1}{w+1}=1$ .