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(B) Given that $\alpha, \beta, \gamma, \delta$ are the roots of $x^4 + x^2 + 1 = 0$. Let $y = x^2$. Since $x$ is a root of the given equation,we have

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$(x^2 + x + 1)^2 = 0$

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(B) Given that $\alpha, \beta, \gamma, \delta$ are the roots of $x^4 + x^2 + 1 = 0$. Let $y = x^2$. Since $x$ is a root of the given equation,we have $x^4 + x^2 + 1 = 0$. Substituting $x^2 = y$,we get $y^2 + y + 1 = 0$. Since the roots of the new equation are $y_1 = \alpha^2, y_2 = \beta^2, y_3 = \gamma^2, y_4 = \delta^2$,and each root of the original equation satisfies $x^4 + x^2 + 1 = 0$,we observe that the equation $y^2 + y + 1 = 0$ is satisfied by these squared values. Since the original equation is of degree $4$,it has $4$ roots. The equation $y^2 + y + 1 = 0$ is of degree $2$,but it represents the squared values. Specifically,$x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1) = 0$. The roots of $x^2 + x + 1 = 0$ are $\omega, \omega^2$ and the roots of $x^2 - x + 1 = 0$ are $-\omega, -\omega^2$. Squaring these,we get $\omega^2, \omega^4 = \omega$ and $(-\omega)^2 = \omega^2, (-\omega^2)^2 = \omega^4 = \omega$. Thus,the roots $\alpha^2, \beta^2, \gamma^2, \delta^2$ are $\omega, \omega^2, \omega, \omega^2$. These are the roots of $(x^2 + x + 1)^2 = 0$.

Let $\alpha, \beta, \gamma, \delta$ be the roots of the equation $x^4 + x^2 + 1 = 0$. Then the equation whose roots are $\alpha^2, \beta^2, \gamma^2, \delta^2$ is:

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