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(A) The given equation is $x^4+x^2+1=0$. Since the equation contains only even powers of $x$,if $x$ is a root,then $-x$ is also a root. Let the roots

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(A) The given equation is $x^4+x^2+1=0$. Since the equation contains only even powers of $x$,if $x$ is a root,then $-x$ is also a root. Let the roots be $\alpha, \beta, \gamma, \delta$. Since the roots occur in pairs of $\pm x_i$,we can write them as $\alpha, -\alpha, \gamma, -\gamma$. For any odd power $n$,the sum of the $n$-th powers of the roots is $\alpha^n + (-\alpha)^n + \gamma^n + (-\gamma)^n$. If $n$ is odd,$\alpha^n + (-\alpha)^n = \alpha^n - \alpha^n = 0$. Thus,$\alpha^3+\beta^3+\gamma^3+\delta^3 = 0$. Since the numerator is $0$ and the denominator $\alpha^6+\beta^6+\gamma^6+\delta^6$ is non-zero,the value of the expression is $0$.

If $\alpha, \beta, \gamma, \delta$ are the roots of the equation $x^4+x^2+1=0$,then $\frac{\alpha^3+\beta^3+\gamma^3+\delta^3}{\alpha^6+\beta^6+\gamma^6+\delta^6}=$

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