Let x_1, x_2, x_3 in mathbb{R} setminus {0},x | vedclass

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(B) Given $\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} = \frac{1}{x_1 + x_2 + x_3}$.This simplifies to $\frac{x_1 x_2 + x_2 x_3 + x_3 x_1}{x_1 x_2 x

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all odd integers $n$

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(B) Given $\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} = \frac{1}{x_1 + x_2 + x_3}$.This simplifies to $\frac{x_1 x_2 + x_2 x_3 + x_3 x_1}{x_1 x_2 x_3} = \frac{1}{x_1 + x_2 + x_3}$.$(x_1 + x_2 + x_3)(x_1 x_2 + x_2 x_3 + x_3 x_1) = x_1 x_2 x_3$.Let $x_1, x_2, x_3$ be roots of the cubic equation $t^3 + \alpha t^2 + \beta t + \gamma = 0$,where $\alpha = -(x_1 + x_2 + x_3)$,$\beta = x_1 x_2 + x_2 x_3 + x_3 x_1$,and $\gamma = -x_1 x_2 x_3$.The equation becomes $(-\alpha)(\beta) = -(-\gamma) \implies \gamma = \alpha \beta$.Substituting this into the cubic equation: $t^3 + \alpha t^2 + \beta t + \alpha \beta = 0$.$t^2(t + \alpha) + \beta(t + \alpha) = 0 \implies (t + \alpha)(t^2 + \beta) = 0$.Thus,the roots are $t = -\alpha, \sqrt{-\beta}, -\sqrt{-\beta}$.Let $x_1 = -\alpha$,$x_2 = k$,and $x_3 = -k$ where $k = \sqrt{-\beta}$.Then $x_1^n + x_2^n + x_3^n = x_1^n + k^n + (-k)^n$.If $n$ is an odd integer,$k^n + (-k)^n = 0$,so $x_1^n + x_2^n + x_3^n = x_1^n$.Also,$\frac{1}{x_1^n} + \frac{1}{x_2^n} + \frac{1}{x_3^n} = \frac{1}{x_1^n} + \frac{1}{k^n} + \frac{1}{(-k)^n} = \frac{1}{x_1^n} + 0 = \frac{1}{x_1^n}$.Thus,the equality holds for all odd integers $n$.

Let $x_1, x_2, x_3 \in \mathbb{R} \setminus \{0\}$,$x_1 + x_2 + x_3 \neq 0$ and $\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} = \frac{1}{x_1 + x_2 + x_3}$. Then $\frac{1}{x_1^n + x_2^n + x_3^n} = \frac{1}{x_1^n} + \frac{1}{x_2^n} + \frac{1}{x_3^n}$ holds good for:

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