Question

If 1, $\omega$ and ${\omega }^2$ are the cube roots of unity, evaluate $(1-{\omega }^4+{\omega }^8)(1-{\omega }^8+{\omega }^{16})$

Answer 1

$(1-{\omega }^4+{\omega }^8)(1-{\omega }^8+{\omega }^{16})$
${\omega }^4={\omega }^3.\omega =\omega ,{\omega }^8={\omega }^6.{\omega }^2={\omega }^2$ $(\because {\omega }^3=1)$
${\omega }^{16}={\omega }^{15}.\omega =\omega$
$\therefore (1-{\omega }^4+{\omega }^8)(1-{\omega }^8+{\omega }^{16})$
$=(1-\omega +{\omega }^2)(1-{\omega }^2+\omega )(\because 1+\omega +{\omega }^2=0)$
$=(-2\omega )(-2{\omega }^2)$
$=4{\omega }^3$
$=4.1=4$