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If $α, β, γ$ are the roots of $x^{3} - 3 x + 1 = 0$ then the equation whose roots are
$α - \frac{1}{β γ}$ , $β - \frac{1}{γ α}$ , $γ - \frac{1}{α β}$ is

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$x^{3} - 12 x + 8 = 0$

$x^{3} - 3 x + 8 = 0$

$x^{3} - 6 x + 8 = 0$

$x^{3} - 9 x + 8 = 0$

answer is D.

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Detailed Solution

$α + β + γ = 0, αβ + βγ + γ = - 3 αβγ = - 1$

now $y = \frac{α β γ - 1}{β γ} = \frac{- 2}{β γ} = \frac{- 2 α}{- 1} = 2 α$

$α = \frac{y}{2}$

$f (\frac{y}{2}) = 0 \Rightarrow y^{3} - 12 y + 8 = 0$

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