Common roots of the equations z^3 + 2z^2 + 2z | vedclass

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Question

(A) The first equation can be factored as $(z + 1)(z^2 + z + 1) = 0$.Its roots are $-1, \omega, 	ext{ and } \omega^2$.Let $f(z) = z^{1985} + z^{100}

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$\omega, \omega^2$

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(A) The first equation can be factored as $(z + 1)(z^2 + z + 1) = 0$.Its roots are $-1, \omega, 	ext{ and } \omega^2$.Let $f(z) = z^{1985} + z^{100} + 1$.Testing $z = -1$: $f(-1) = (-1)^{1985} + (-1)^{100} + 1 = -1 + 1 + 1 = 1 
eq 0$.Therefore,$-1$ is not a root of $f(z) = 0$.Testing $z = \omega$: $f(\omega) = \omega^{1985} + \omega^{100} + 1$.Since $\omega^3 = 1$,we have $f(\omega) = (\omega^3)^{661} \cdot \omega^2 + (\omega^3)^{33} \cdot \omega + 1 = \omega^2 + \omega + 1 = 0$.Therefore,$\omega$ is a root of $f(z) = 0$.Testing $z = \omega^2$: $f(\omega^2) = (\omega^2)^{1985} + (\omega^2)^{100} + 1 = \omega^{3970} + \omega^{200} + 1$.$f(\omega^2) = (\omega^3)^{1323} \cdot \omega + (\omega^3)^{66} \cdot \omega^2 + 1 = \omega + \omega^2 + 1 = 0$.Therefore,$\omega^2$ is also a root of $f(z) = 0$.Thus,the common roots are $\omega 	ext{ and } \omega^2$.

Common roots of the equations $z^3 + 2z^2 + 2z + 1 = 0$ and $z^{1985} + z^{100} + 1 = 0$ are

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