Question

The common roots of the equations $z^{3}+2 z^{2}+2 z+1=0$ and $z^{2018}+z^{2017}+1=0$ satisfy the equation

• A. $z^{2}-z+1=0$
• B. $z^{4}+z^{2}+1=0$
• C. $z^{6}+z^{3}+1=0$
• D. $z^{12}+z^{6}-1=0$

Answer 1

Answer: (B)

We have,
$ z^{3}+2 z^{2}+2 z+1=0 $
$ \Rightarrow \, z^{3}+1+2 z(z+1)=0 $
$ \Rightarrow (z+1)\left(z^{2}-z+1\right)+2 z(z+1)=0 $
$ \Rightarrow \, (z+1)\left(z^{2}-z+1+2 z\right)=0 $
$\Rightarrow \,(z+1)\left(z^{2}+z+1\right)=0 $
So, $ z+1=0$ and $z^{2}+z+1=0$
$z=-1 \Rightarrow z=\omega, \omega^{2}$
Hence, roots of $z^{3}+2 z^{2}+2 z+1$ are $-1, \omega, \omega^{2}$
for $z=-1$
$z^{2018}+z^{2017}+1=(-1)^{2018}+(-1)^{2017}+1$
$=+1-1+1=1 \neq 0$
for $\quad z=w$
$z^{2018}+z^{2017}+1=(\omega)^{2018}+(\omega)^{2017}+1=\omega^{2}+\omega+1$
$=0 \, \left[\because \omega^{2}+\omega+1=0\right]$
for $z=\omega^{2}$
$z^{2018}+z^{2017}+1=\left(\omega^{2}\right)^{2018}+\left(\omega^{2}\right)^{2017}+1$
$=\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 $z =\omega^{2}$
$\left(\omega^{2}\right)^{4}+\left(\omega^{2}\right)^{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.