Answer:
Option B
Explanation:
We have
$z^{3}+2z^{3}+2z+1=0$
$\Rightarrow$ $z^{3}+1+2z(z+1)$=0
$\Rightarrow$ $(z+1)(z^{2}-z+1)+2z(z+1)=0$
$\Rightarrow$ $(z+1)(z^{2}-z+1+2z)=0$
$\Rightarrow$ $(z+1)(z^{2}+z+1)=0$
So, z+1=0 and $z^{2}+z+1=0$
z=-1$\Rightarrow$ 4z= $\omega,\omega^{2}$
hence, roots of $z^{3}+2z^{2}+2z+1$ are $-1, \omega, \omega^{2}$ for z=-1
$z^{2018}+z^{2017}+1= (-1)^{2018}+(-1)^{2017}+1$
=+1-1+1=1≠0
for z=w
$z^{2018}+z^{2017}+1=(\omega)^{2018}+(\omega)^{2017}+1=\omega^{2}+\omega+1$= 0
[$\because \omega^{2}+\omega+1=0]$
for $z=\omega^{2}$
$z^{2018}+z^{2017}+1=$ $ (\omega^{2})^{2018}+(\omega^{2})^{2017}+1$
= $\omega^{4036}+\omega^{4034}+1$
=$\omega+\omega^{2}+1=0$
Thus, the common roots are $\omega$ and $\omega^{2}$ by checking options
$z^{4}+z^{2}+1=0$
for $z=\omega$
$\omega^{4}+\omega^{2}+1=\omega+\omega^{2}+1=0$
and for $z=\omega^{2}$
$(\omega^{2})^{4}+(\omega^{2})^{2}+1=\omega^{8}+\omega^{4}+1$
= $\omega^{2}+\omega+1=0$
Hence, $z^{4}+z^{2}+1=0$ satisfy by the both common roots
