Question

if w is complex cube root of unity then( 1+ w+ w square)7 = 2
true and false ​

Answer 1

The three cube roots of unity are numbers that satisfy $\rm x^{3}=1$.

$\rm x^{3}=1$

$\;$

$\rm x^{3}-1=0$

$\;$

$\rm(x-1)(x^{2}+x+1)=0$

$\;$

$\rm\therefore x=1\ or\ x=\omega\ or\ x=\bar{\omega}$

$\;$

The complex cube root of unity then satisfies $\rm\omega^{2}+\omega+1=0$, because the discriminant of the quadratic factor is negative.

$\rm(\omega^{2}+\omega+1)^{7}=0^{7}$

$\;$

$\rm\therefore(\omega^{2}+\omega+1)^{7}=0$

Hence, the given equation is false.