If $\alpha,\beta,\gamma$ are the roots of the equation $x^{3} + 4x - 1 = 0$, then $(\alpha + \beta)^{- 1} + (\beta + \gamma)^{- 1} + (\gamma + \alpha)^{- 1}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(c)
Here, $\alpha + \beta + \gamma = 0,\ \ \ \alpha\beta + \beta\gamma + \gamma\alpha = 4$
And $\alpha\beta\gamma = 1$
$	herefore\ \frac{1}{\alpha + \beta} + \frac{1}{\beta + \gamma} + \frac{1}{\gamma + \alpha} = - \frac{1}{\gamma} - \frac{1}{\alpha} - \frac{1}{\beta}$
$= - \left\lbrack \frac{1}{\gamma} + \frac{1}{\alpha} + \frac{1}{\beta} ightbrack = - \left\lbrack \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma} ightbrack = - 4$

Q. If $\alpha,\beta,\gamma$ are the roots of the equation $x^{3} + 4x - 1 = 0$, then $(\alpha + \beta)^{- 1} + (\beta + \gamma)^{- 1} + (\gamma + \alpha)^{- 1}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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