Let x_1, x_2, x_3 in R - {0},x_1 + x_2 + x_3 | 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 numbers $n$

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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 $t^3 + \alpha t^2 + \beta t + \gamma = 0$,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 $(-\alpha)(\beta) = -(-\gamma) \Rightarrow \gamma = \alpha \beta$.So,$t^3 + \alpha t^2 + \beta t + \alpha \beta = 0 \Rightarrow t^2(t + \alpha) + \beta(t + \alpha) = 0 \Rightarrow (t + \alpha)(t^2 + \beta) = 0$.The roots are $t_1 = -\alpha$,$t_2 = \sqrt{-\beta}$,$t_3 = -\sqrt{-\beta}$.Thus,$x_1 = -(x_1+x_2+x_3) \Rightarrow x_2 + x_3 = 0 \Rightarrow x_3 = -x_2$.Now,check the condition $\frac{1}{x_1^n + x_2^n + x_3^n} = \frac{1}{x_1^n} + \frac{1}{x_2^n} + \frac{1}{x_3^n}$.For $x_3 = -x_2$,the expression becomes $\frac{1}{x_1^n + x_2^n + (-x_2)^n} = \frac{1}{x_1^n} + \frac{1}{x_2^n} + \frac{1}{(-x_2)^n}$.If $n$ is odd,$(-x_2)^n = -x_2^n$,so $\frac{1}{x_1^n + x_2^n - x_2^n} = \frac{1}{x_1^n} + \frac{1}{x_2^n} - \frac{1}{x_2^n} \Rightarrow \frac{1}{x_1^n} = \frac{1}{x_1^n}$,which is true.

Let $x_1, x_2, x_3 \in R - \{0\}$,$x_1 + x_2 + x_3 \neq 0$ 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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