If alpha, beta and gamma are the zeroes of the cubic polynomial x3+4x+2, then find the value of (1/alpha+beta) + (1/beta+gamma) + (1/alpha+gamma).

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Solution

$Let p (x) = x^{3} + 4 x + 2 Now, α + β + γ = - \frac{coefficient of x^{2}}{coefficient of x^{3}} = - \frac{0}{1} = 0 Now, αβ + βγ + γα = \frac{coefficient of x}{coefficient of x^{3}} = 4 Now, αβγ = - \frac{constant term}{coefficient of x^{3}} = - 2 Now, α + β + γ = 0 \Rightarrow α + β = - γ; β + γ = - α; γ + α = - β Now, \frac{1}{α + β} + \frac{1}{β + γ} + \frac{1}{γ + α} = - \frac{1}{γ} - \frac{1}{α} - \frac{1}{β} = \frac{- αβ - βγ - γα}{αβγ} = - \frac{(αβ + βγ + γα)}{αβγ} = - \frac{4}{(- 2)} = 2$

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