Question

The common roots of the equations $z^3+2z^2+2z+1=0$ 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,
$z^3+2z^2+2z+1=0$
$\Rightarrow z^3+1+2z(z+1)=0$
$\Rightarrow (z+1)\left(z^2-z+1\right)+2z(z+1)=0$
$\Rightarrow (z+1)\left(z^2-z+1+2z\right)=0$
$\Rightarrow (z+1)\left(z^2+z+1\right)=0$
So, $z+1=0$ and $z^2+z+1=0$
$z=-1\Rightarrow z=\omega ,{\omega }^2$
Hence, roots of $z^3+2z^2+2z+1$ are $-1,\omega ,{\omega }^2$
for $z=-1$
$z^{2018}+z^{2017}+1=(-1{)}^{2018}+(-1{)}^{2017}+1$
$=+1-1+1=1\ne 0$
for $z=w$
$z^{2018}+z^{2017}+1=(\omega {)}^{2018}+(\omega {)}^{2017}+1={\omega }^2+\omega +1$
$=0\left[\because {\omega }^2+\omega +1=0\right]$
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.