Question

The common roots of the equations $z^{3}+2 z^{2}+2 z+1=0, z^{2014}+z^{2015}+1=0$ are

• A. $\omega, \omega^{2}$
• B. $1, \omega, \omega^{2}$
• C. $-1, \omega, \omega^{2}$
• D. $-\omega,-\omega^{2}$

Answer 1

The Correct Option is A

The given equation
$z^{3}+2 z^{2}+2 z+1=0$ can be rewritten as
$(z+1)\left(z^{2}+z+1\right)=0$
Since, its roots are $-1, \omega$ and $\omega^{2}$.
Let $f(z)=z^{2014}+z^{2015}+1=0$
Put $z=-1, \omega$ and $\omega^{2}$ respectively, we get
$f(-1) =(-1)^{2014}+(-1)^{2015}+1=0$
$=1 \neq 0$
Therefore, $-1$ is not a root of the equation $f(z)=0$.
Again,$ f(\omega) =(\omega)^{2014}+(\omega)^{2015}+1=0 $
$=\left(\omega^{3}\right)^{671} \cdot \omega+\left(\omega^{3}\right)^{671} \cdot \omega^{2}+1=0 $
$ \Rightarrow \omega+\omega^{2}+1=0 $
$\Rightarrow \omega^{2}+\omega+1=0 $
$\Rightarrow 0=0$
Therefore, $\omega$ is a root of the equation $f(z)=0$
Similarly, $f\left(\omega^{3}\right)=\left(\omega^{2}\right)^{2014}+\left(\omega^{3}\right)^{2015}+1=0 $
$\Rightarrow \left(\omega^{3}\right)^{1342} \cdot m^{2}+\left(\omega^{3}\right)^{1343} \cdot m+1=0 $
$ \Rightarrow \omega^{2}+\omega+1=0 $
$ \Rightarrow 0=0 $
Hence, $\omega$ and $\omega^{2}$ are the common roots.