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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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Given $\alpha, \beta, \gamma, \delta$ are roots of $x^4+x^2+1=0$
$\therefore \quad x^2=\frac{-1 \pm \sqrt{1-4}}{2}=\frac{-1 \pm \sqrt{3} i}{2}$
$\therefore \quad x= \pm \sqrt{\frac{1+\sqrt{3} i}{2}}, x= \pm \sqrt{\frac{-1-\sqrt{3} i}{2}}$
Let $\alpha=\sqrt{\frac{1+\sqrt{3} i}{2}}, \beta=-\sqrt{\frac{1+\sqrt{3} i}{2}}, \gamma=\sqrt{\frac{-1-\sqrt{3} i}{2}}$ $\delta=-\sqrt{\frac{-1-\sqrt{3} i}{2}}$
$\Rightarrow \quad \alpha^3+\beta^3+\gamma^3+\delta^3=0$
$\therefore$ Given expression, $\frac{\alpha^3+\beta^3+\gamma^3+\delta^3}{\alpha^6+\beta^6+\gamma^6+\delta^6}=0$
$\therefore \quad x^2=\frac{-1 \pm \sqrt{1-4}}{2}=\frac{-1 \pm \sqrt{3} i}{2}$
$\therefore \quad x= \pm \sqrt{\frac{1+\sqrt{3} i}{2}}, x= \pm \sqrt{\frac{-1-\sqrt{3} i}{2}}$
Let $\alpha=\sqrt{\frac{1+\sqrt{3} i}{2}}, \beta=-\sqrt{\frac{1+\sqrt{3} i}{2}}, \gamma=\sqrt{\frac{-1-\sqrt{3} i}{2}}$ $\delta=-\sqrt{\frac{-1-\sqrt{3} i}{2}}$
$\Rightarrow \quad \alpha^3+\beta^3+\gamma^3+\delta^3=0$
$\therefore$ Given expression, $\frac{\alpha^3+\beta^3+\gamma^3+\delta^3}{\alpha^6+\beta^6+\gamma^6+\delta^6}=0$
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