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(C) Since $\alpha, \beta, \gamma$ are the roots of the equation $x^3+x+10=0$,we have $\alpha+\beta+\gamma=0$.Now,$\alpha_1=\frac{\alpha+\beta}{\gamma^

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$\frac{3}{10}$

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(C) Since $\alpha, \beta, \gamma$ are the roots of the equation $x^3+x+10=0$,we have $\alpha+\beta+\gamma=0$.Now,$\alpha_1=\frac{\alpha+\beta}{\gamma^2}=\frac{-\gamma}{\gamma^2}=\frac{-1}{\gamma}$.Similarly,$\beta_1=\frac{-1}{\alpha}$ and $\gamma_1=\frac{-1}{\beta}$.Thus,$\alpha_1, \beta_1, \gamma_1$ are the roots of the equation obtained by substituting $x = -\frac{1}{y}$ in $x^3+x+10=0$.$(-\frac{1}{y})^3 + (-\frac{1}{y}) + 10 = 0 \implies -\frac{1}{y^3} - \frac{1}{y} + 10 = 0$.Multiplying by $-y^3$,we get $10y^3 - y^2 - 1 = 0$.Since $\alpha_1, \beta_1, \gamma_1$ are roots of $10x^3-x^2-1=0$,we have $10\alpha_1^3-\alpha_1^2-1=0$,$10\beta_1^3-\beta_1^2-1=0$,and $10\gamma_1^3-\gamma_1^2-1=0$.Summing these equations: $10(\alpha_1^3+\beta_1^3+\gamma_1^3) - (\alpha_1^2+\beta_1^2+\gamma_1^2) - 3 = 0$.Dividing by $10$: $(\alpha_1^3+\beta_1^3+\gamma_1^3) - \frac{1}{10}(\alpha_1^2+\beta_1^2+\gamma_1^2) = \frac{3}{10}$.

Let $\alpha, \beta, \gamma$ be the roots of $x^3+x+10=0$ and $\alpha_1=\frac{\alpha+\beta}{\gamma^2}, \beta_1=\frac{\beta+\gamma}{\alpha^2}, \gamma_1=\frac{\gamma+\alpha}{\beta^2}$. Then,the value of $(\alpha_1^3+\beta_1^3+\gamma_1^3)-\frac{1}{10}(\alpha_1^2+\beta_1^2+\gamma_1^2)$ is

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