If alpha, beta and gamma are roots of x^{3}-2 | vedclass

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(B) Given the cubic equation $x^{3}-2x+1=0$ with roots $\alpha, \beta, \gamma$.From Vieta's formulas,we have:$\alpha+\beta+\gamma = 0$$\alpha\beta+\be

Vedclass · Accepted Answer

$-1$

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(B) Given the cubic equation $x^{3}-2x+1=0$ with roots $\alpha, \beta, \gamma$.From Vieta's formulas,we have:$\alpha+\beta+\gamma = 0$$\alpha\beta+\beta\gamma+\gamma\alpha = -2$$\alpha\beta\gamma = -1$We need to evaluate $\sum\left(\frac{1}{\alpha+\beta-\gamma}ight)$.Since $\alpha+\beta+\gamma = 0$,we have $\alpha+\beta = -\gamma$.Substituting this into the expression:$\sum\left(\frac{1}{-\gamma-\gamma}ight) = \sum\left(\frac{1}{-2\gamma}ight) = -\frac{1}{2} \sum\left(\frac{1}{\gamma}ight)$$= -\frac{1}{2} \left(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}ight)$$= -\frac{1}{2} \left(\frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma}ight)$$= -\frac{1}{2} \left(\frac{-2}{-1}ight) = -\frac{1}{2} 	imes 2 = -1$.

If $\alpha, \beta$ and $\gamma$ are roots of $x^{3}-2x+1=0$,then the value of $\sum\left(\frac{1}{\alpha+\beta-\gamma}\right)$ is

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