Let x_1,x_2,x_3 in R-{0} ,x_1 + x_2 + x | vedclass

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Question

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}^

Vedclass · Accepted Answer

all odd numbers $n$

Vedclass · Answer

Consider a cubic equation whose roots are

$\mathrm{x}_{1}, \mathrm{x}_{2}$ and $\mathrm{x}_{3}$

$f(\mathrm{x})=\mathrm{x}^{3}+\alpha \mathrm{x}^{2}+\beta \mathrm{x}+\gamma$

$=\left(\mathrm{x}-\mathrm{x}_{1}ight)\left(\mathrm{x}-\mathrm{x}_{2}ight)\left(\mathrm{x}-\mathrm{x}_{3}ight)=0$

Given $\left(\mathrm{x}_{1} \mathrm{x}_{2}+\mathrm{x}_{2} \mathrm{x}_{3}+\mathrm{x}_{3} \mathrm{x}_{1}ight)\left(\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}ight)=\mathrm{x}_{1} \mathrm{x}_{2} \mathrm{x}_{3}$

$\beta(-\alpha)=-\gamma$

$\gamma=\alpha \beta$

$	herefore f(\mathrm{x})=0=\mathrm{x}^{3}+\alpha \mathrm{x}^{2}+\beta \mathrm{x}+\alpha \beta$

$=\mathrm{x}^{2}(\mathrm{x}+\alpha)+\beta(\mathrm{x}+\alpha)=(\mathrm{x}+\alpha)\left(\mathrm{x}^{2}+\betaight)$

$	herefore \quad \mathrm{x}=-\alpha$

Let $\mathrm{x}_{1}=-\alpha ; \mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}=-\alpha \Rightarrow \mathrm{x}_{2}+\mathrm{x}_{3}=0$

$\Rightarrow \quad x_{2}=-x_{3}$

$	herefore \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}ight)^{\mathrm{n}}+\left(-\mathrm{x}_{2}ight)^{\mathrm{n}}}$

$=\frac{1}{x_{1}^{n}}+\frac{1}{x_{2}^{n}}+\frac{1}{\left(-x_{2}ight)^{n}}$

Which is true for all odd integer $n$.

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

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