The common roots of the equations z^3+2z^2+2z | vedclass

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(B) We have,$z^3+2z^2+2z+1=0$.Factoring the expression: $(z^3+1)+2z(z+1)=0$.$(z+1)(z^2-z+1)+2z(z+1)=0$.$(z+1)(z^2-z+1+2z)=0$.$(z+1)(z^2+z+1)=0$.The ro

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$z^4+z^2+1=0$

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(B) We have,$z^3+2z^2+2z+1=0$.Factoring the expression: $(z^3+1)+2z(z+1)=0$.$(z+1)(z^2-z+1)+2z(z+1)=0$.$(z+1)(z^2-z+1+2z)=0$.$(z+1)(z^2+z+1)=0$.The roots are $z=-1$ and the roots of $z^2+z+1=0$,which are $\omega$ and $\omega^2$.Now,test these roots in $z^{2018}+z^{2017}+1=0$:For $z=-1$: $(-1)^{2018}+(-1)^{2017}+1 = 1-1+1 = 1 
eq 0$.For $z=\omega$: $\omega^{2018}+\omega^{2017}+1 = (\omega^3)^{672} \cdot \omega^2 + (\omega^3)^{672} \cdot \omega + 1 = \omega^2+\omega+1 = 0$.For $z=\omega^2$: $(\omega^2)^{2018}+(\omega^2)^{2017}+1 = \omega^{4036}+\omega^{4034}+1 = \omega+\omega^2+1 = 0$.The common roots are $\omega$ and $\omega^2$.Checking the options for $z=\omega$ and $z=\omega^2$:For $z^4+z^2+1=0$:If $z=\omega$,$\omega^4+\omega^2+1 = \omega+\omega^2+1 = 0$.If $z=\omega^2$,$(\omega^2)^4+(\omega^2)^2+1 = \omega^8+\omega^4+1 = \omega^2+\omega+1 = 0$.Thus,the common roots satisfy $z^4+z^2+1=0$.

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

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