If alpha, beta, gamma, delta are the roots of | vedclass

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(A) The given equation is $x^4+x^2+1=0$. This can be factored as $(x^2+x+1)(x^2-x+1)=0$. The roots of $x^2+x+1=0$ are $\omega$ and $\omega^2$,where $\

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(A) The given equation is $x^4+x^2+1=0$. This can be factored as $(x^2+x+1)(x^2-x+1)=0$. The roots of $x^2+x+1=0$ are $\omega$ and $\omega^2$,where $\omega$ is a complex cube root of unity. The roots of $x^2-x+1=0$ are $-\omega$ and $-\omega^2$. Given $\alpha+\beta=-1$ and $\alpha^2=\beta$,we identify $\alpha=\omega$ and $\beta=\omega^2$ (since $\omega+\omega^2=-1$ and $\omega^2=\omega^2$). Given $\gamma+\delta=1$ and $\gamma^2=-\delta$,we identify $\gamma=-\omega$ and $\delta=-\omega^2$ (since $-\omega-\omega^2=1$ and $(-\omega)^2 = \omega^2 = -(-\omega^2)$). Now,calculate $\alpha^{2023}+\beta^{2023}+\gamma^{2022}+\delta^{2022}$: $= \omega^{2023} + (\omega^2)^{2023} + (-\omega)^{2022} + (-\omega^2)^{2022}$ $= \omega^{2023} + \omega^{4046} + \omega^{2022} + \omega^{4044}$ Using $\omega^3=1$: $= \omega^1 + \omega^1 + 1 + 1 = 2\omega + 2$. Wait,re-evaluating the sum: $\omega^{2023} = \omega$,$\omega^{4046} = \omega^2$,$\omega^{2022} = 1$,$\omega^{4044} = 1$. Sum $= \omega + \omega^2 + 1 + 1 = (-1) + 2 = 1$.

If $\alpha, \beta, \gamma, \delta$ are the roots of the equation $x^4+x^2+1=0$ such that $\alpha+\beta=-1, \gamma+\delta=1, \alpha^2=\beta$ and $\gamma^2=-\delta$,then $\alpha^{2023}+\beta^{2023}+\gamma^{2022}+\delta^{2022}=$

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