Question

Let $z=\frac{1-i\sqrt{3}}{2},i=\sqrt{-1}.$ Then the value of $21+{(z+\frac{1}{z})}^3+{(z^2+\frac{1}{z^2})}^3+{(z^3+\frac{1}{z^3})}^3+\cdots +{(z^{21}+\frac{1}{z^{21}})}^3$ is

Answer 1

$z=\frac{1-\sqrt{3}i}{2}$
$\therefore z=-\left[\frac{-1+\sqrt{3}i}{2}\right]=-\omega$

$21+{(z+\frac{1}{z})}^3+{(z^2+\frac{1}{z^2})}^3+{(z^3+\frac{1}{z^3})}^3+\cdots +{(z^{21}+\frac{1}{z^{21}})}^3$
$=21+{(-\omega -\frac{1}{\omega })}^3+{\left(\right(-\omega {)}^2+{(-\frac{1}{\omega })}^2)}^3+{\left(\right(-\omega {)}^3+{(-\frac{1}{{\omega }^3})}^3)}^3+\cdots +{\left(\right(-\omega {)}^{21}+\frac{1}{(-\omega {)}^{21}})}^3$
$=21-(\omega +{\omega }^2{)}^3+(\omega +{\omega }^2{)}^3+(-1-1{)}^3+(\omega +{\omega }^2{)}^3-(\omega +{\omega }^2{)}^3+(1+1{)}^3+\cdots +(-\omega -{\omega }^2{)}^3+({\omega }^2+\omega {)}^3+(-{\omega }^3-{\omega }^3{)}^3$
$=21+(1-1-8)+(-1+1+8)+(1-1-8)+(-1+1+8)+(1-1-8)+(-1+1+8)+(1-1-8)$
$=21-8=13$