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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 $\frac{1}{{x^n}_1+{x^n}_2+{x^n}_3} =\frac{1}{{x^n}_1}+\frac{1}{{x^n}_2}+\frac{1}{{x^n}_3}$ holds good for
all $n \in N$
all odd numbers $n$
all even number $n$
no integer $n$
Solution
Consider a cubic equation whose roots are
$\mathrm{x}_{1}, \mathrm{x}_{2}$ and $\mathrm{x}_{3}$
$f(\mathrm{x})=\mathrm{x}^{3}+\alpha \mathrm{x}^{2}+\beta \mathrm{x}+\gamma$
$=\left(\mathrm{x}-\mathrm{x}_{1}\right)\left(\mathrm{x}-\mathrm{x}_{2}\right)\left(\mathrm{x}-\mathrm{x}_{3}\right)=0$
Given $\left(\mathrm{x}_{1} \mathrm{x}_{2}+\mathrm{x}_{2} \mathrm{x}_{3}+\mathrm{x}_{3} \mathrm{x}_{1}\right)\left(\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}\right)=\mathrm{x}_{1} \mathrm{x}_{2} \mathrm{x}_{3}$
$\beta(-\alpha)=-\gamma$
$\gamma=\alpha \beta$
$\therefore f(\mathrm{x})=0=\mathrm{x}^{3}+\alpha \mathrm{x}^{2}+\beta \mathrm{x}+\alpha \beta$
$=\mathrm{x}^{2}(\mathrm{x}+\alpha)+\beta(\mathrm{x}+\alpha)=(\mathrm{x}+\alpha)\left(\mathrm{x}^{2}+\beta\right)$
$\therefore \quad \mathrm{x}=-\alpha$
Let $\mathrm{x}_{1}=-\alpha ; \mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}=-\alpha \Rightarrow \mathrm{x}_{2}+\mathrm{x}_{3}=0$
$\Rightarrow \quad x_{2}=-x_{3}$
$\therefore \frac{1}{\mathrm{x}_{1}^{\mathrm{n}}+\mathrm{x}_{2}^{\mathrm{n}}+\mathrm{x}_{3}^{\mathrm{n}}}=\frac{1}{\mathrm{x}_{1}^{\mathrm{n}}+\left(\mathrm{x}_{2}\right)^{\mathrm{n}}+\left(-\mathrm{x}_{2}\right)^{\mathrm{n}}}$
$=\frac{1}{x_{1}^{n}}+\frac{1}{x_{2}^{n}}+\frac{1}{\left(-x_{2}\right)^{n}}$
Which is true for all odd integer $n$.
