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(D) Given that $\alpha_1, \beta_1, \gamma_1, \delta_1$ are the roots of $a x^4+b x^3+c x^2+d x+e=0$. The roots of $e x^4+d x^3+c x^2+b x+a=0$ are the

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$8$

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(D) Given that $\alpha_1, \beta_1, \gamma_1, \delta_1$ are the roots of $a x^4+b x^3+c x^2+d x+e=0$. The roots of $e x^4+d x^3+c x^2+b x+a=0$ are the reciprocals of the roots of the first equation. Thus,$\alpha_2 = \frac{1}{\delta_1}, \beta_2 = \frac{1}{\gamma_1}, \gamma_2 = \frac{1}{\beta_1}, \delta_2 = \frac{1}{\alpha_1}$. Given $\alpha_1 - \delta_2 = 2 \implies \alpha_1 - \frac{1}{\alpha_1} = 2 \implies \alpha_1^2 - 2\alpha_1 - 1 = 0$. Given $\delta_1 - \alpha_2 = 4 \implies \delta_1 - \frac{1}{\delta_1} = 4 \implies \delta_1^2 - 4\delta_1 - 1 = 0$. Since $\alpha_1$ and $\delta_1$ are roots of the quartic equation,the quadratic factors are $(x^2 - 2x - 1)$ and $(x^2 - 4x - 1)$. Thus,$a x^4+b x^3+c x^2+d x+e = (x^2 - 2x - 1)(x^2 - 4x - 1)$. To find $a+b+c+d+e$,we substitute $x=1$: $a(1)^4 + b(1)^3 + c(1)^2 + d(1) + e = (1^2 - 2(1) - 1)(1^2 - 4(1) - 1)$. $a+b+c+d+e = (-2)(-4) = 8$.

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