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(D) Let the roots of the equation $x^3+x+1=0$ be $\alpha, \beta, \gamma$. Let $S_n = \alpha^n + \beta^n + \gamma^n$.By Newton's Sums for the equation

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$2$

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(D) Let the roots of the equation $x^3+x+1=0$ be $\alpha, \beta, \gamma$. Let $S_n = \alpha^n + \beta^n + \gamma^n$.By Newton's Sums for the equation $x^3+p_1 x^2+p_2 x+p_3=0$,where $p_1=0, p_2=1, p_3=1$,we have the relation $S_n + p_1 S_{n-1} + p_2 S_{n-2} + p_3 S_{n-3} = 0$.For $n=1$: $S_1 + 0 = 0 \Rightarrow S_1 = 0$.For $n=2$: $S_2 + 0 \cdot S_1 + 1 \cdot S_0 + 2 \cdot 1 = 0$ (where $S_0 = 3$ for a cubic equation). Thus,$S_2 + 3 = 0 \Rightarrow S_2 = -3$ is incorrect; let's use the recurrence $S_n + S_{n-2} + S_{n-3} = 0$.$S_1 = 0$.$S_2 = -S_0 - S_{-1} = -3 - 0 = -3$ (using $S_0=3, S_{-1}=0$).$S_3 = -S_1 - S_0 = 0 - 3 = -3$.$S_4 = -S_2 - S_1 = -(-3) - 0 = 3$ is incorrect. Let's re-evaluate: $x^3 = -x-1$. Multiplying by $x$,$x^4 = -x^2-x$. Summing over roots: $S_4 = -S_2 - S_1$.Since $S_1 = 0$ and $S_2 = \alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = 0^2 - 2(1) = -2$.Thus,$S_4 = -(-2) - 0 = 2$.

The sum of the fourth powers of the roots of the equation $x^3+x+1=0$ is

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