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The common roots to the equations $z^3+2z^2+2z+1=0\:\:and\:\:z^{149}+z^{100}+1=0$ are

$\begin{array}{1 1} (A) 1,\omega,\omega^2\\ (B) -1,\omega,\omega^2\\ (C) \omega,\omega^2\\ (D) -\omega,-\omega^2 \end{array} $

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  • If $z^2+z+1=0$, then $z=\omega\:or\:\omega^2$
$z^3+2z^2+2z+1=0$
$\Rightarrow\:(z^3+1)+2z(z+1)=0$
$\Rightarrow\:(z+1)(z^2-z+1)+2z(z+1)=0$
$\Rightarrow\:(z+1)(z^2-z+1+2z)=0$
$\Rightarrow\:(z+1)(z^2+z+1)=0$
$\Rightarrow\:z=-1,\omega,\omega^2$
Substitute $z=-1,\omega,\omega^2$ in the eqn. $z^{149}+z^{100}+1=0$
$\omega\:\:and\:\:\omega^2$ satisfy the eqn., But not $-1$
$\therefore$ the common roots are $\omega\:\:and\:\:\omega^2$
answered Aug 2, 2013 by rvidyagovindarajan_1
 

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