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If $α, β, γ$ are the roots of $x^{3} - 6 x - 4 = 0$ then the equation whose roots are $β γ + \frac{1}{α}, γ α + \frac{1}{β}, α β + \frac{1}{γ}$ is

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$4 x^{3} + 30 x^{2} + 125 = 0$

$x^{3} + 15 x^{2} - 120 = 0$

$4 x^{3} + 30 x^{2} - 125 = 0$

$4 x^{3} - 30 x^{2} - 125 = 0$

answer is C.

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Detailed Solution

$α + β + γ = 0, αβ + βγ + γα = - 6, αβγ = 4$

$y = \frac{α β γ + 1}{α} = \frac{5}{α}$

$\Rightarrow x = \frac{5}{y} \Rightarrow f (\frac{5}{y}) = 0$

$\Rightarrow 4 y^{3} + 30 y^{2} - 125 = 0$

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