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(B) Given that $\alpha, \beta, \gamma, \delta$ are the roots of the equation $x^4 + x^2 + 1 = 0$. Let $y = x^2$,which implies $x = \pm \sqrt{y}$. Subs

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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 the equation $x^4 + x^2 + 1 = 0$. Let $y = x^2$,which implies $x = \pm \sqrt{y}$. Substituting $x^2 = y$ into the original equation $x^4 + x^2 + 1 = 0$,we get $(x^2)^2 + x^2 + 1 = 0$. Replacing $x^2$ with $y$,we obtain $y^2 + y + 1 = 0$. Since the roots of the new equation are $\alpha^2, \beta^2, \gamma^2, \delta^2$,and each root of the original equation satisfies $x^4 + x^2 + 1 = 0$,the transformation $y = x^2$ maps the four roots to two values of $y$ that satisfy $y^2 + y + 1 = 0$. Since each root is squared,the resulting equation must account for the multiplicity of the roots. The equation $y^2 + y + 1 = 0$ has two roots,say $y_1$ and $y_2$. Since the original equation is of degree $4$,the new equation must also be of degree $4$. Thus,the required equation is $(y^2 + y + 1)^2 = 0$,which in terms of $x$ is $(x^2 + x + 1)^2 = 0$.

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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