If $α, β, γ$ be the roots of the equation $x (1 + x^{2}) + x^{2} (6 + x) + 2 = 0$ , then the value of $α^{- 1} + β^{- 1} + γ^{- 1}$ , is:

- 3

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\frac{1}{2}

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- \frac{1}{2}

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none of these

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Solution

The correct option is B $- \frac{1}{2}$
Given, $x (1 + x^{2}) + x^{2} (6 + x) + 2 = 0$
$\Rightarrow x + x^{3} + 6 x^{2} + x^{3} + 2 = 0$
$\Rightarrow 2 x^{3} + 6 x^{2} + x + 2 = 0$
As $α, β, γ$ are roots
Therefore, $s_{1} = α + β + γ = - 3$
$s_{2} = α β + β γ + γ α = \frac{1}{2}$
$s_{3} = α β γ = - 1$
Now, $α^{- 1} + β^{- 1} + γ^{- 1} = \frac{1}{α} + \frac{1}{β} + \frac{1}{γ} = \frac{α β + β γ + γ α}{α β γ} = - \frac{1}{2}$

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