Discussion Forum

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

Answer:

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