The common roots of the equations $z^3+2 z^2+2 z+1$ $=0, z^{2014}+z^{2015}+1=0$ are
The common roots of the equations $z^3+2 z^2+2 z+1$ $=0, z^{2014}+z^{2015}+1=0$ are
$\omega, \omega^2$
$1, \omega, \omega^2$
$-1, \omega, \omega^2$
$-\omega,-\omega^2$
Solution
Given equation,
$\begin{aligned}
& z^3+2 z^2+2 z+1=0 \\
& (z+1)\left(z^2+z+1\right)=0
\end{aligned}$
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 $f(z)=0$
$\begin{aligned}
& \text { Again, } f(\omega)=(\omega)^{2014}+(\omega)^{2015}+1 \\
& =\left(\omega^3\right)^{671} \cdot \omega+\left(\omega^3\right)^{671} \cdot \omega^2+1 \\
& =\omega+\omega^2+1 \\
& =\omega^2+\omega+1 \\
& =0
\end{aligned}$
Therefore, $\omega$ is a root of the equation $f(z)=0$
Similarly,
$\begin{aligned}
& f\left(\omega^3\right)=\left(\omega^2\right)^{2014}+\left(\omega^3\right)^{2015}+1 \\
& =\left(\omega^3\right)^{1342} \cdot \omega^2+\left(\omega^3\right)^{1343} \cdot \omega+1 \\
& =\omega^2+\omega+1 \\
& =0
\end{aligned}$
Hence $\omega$ and $\omega^2$ are the common roots of $z^{2014}+z^{2015}+1=0$