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

$\begin{array}{l} I f α, β, γ a r e r o o t s o f x^{3} - 6 x - 4 = 0 T h e n \\ α + β + γ = 0, α β + β γ + γ α = - 6, α β γ = 4 \\ L e t β γ + \frac{1}{α} = y \Rightarrow \frac{α β γ + 1}{α} = y \\ \Rightarrow \frac{4 + 1}{α} = y \Rightarrow α = \frac{5}{y} \\ ∵ α = \frac{5}{y} i s r o o t o f x^{3} - 6 x - 4 = 0 \\ \Rightarrow {(\frac{5}{y})}^{3} - 6 (\frac{5}{y}) - 4 = 0 \\ \Rightarrow \frac{125}{y^{3}} - \frac{30}{y} - 4 = 0 \\ \Rightarrow 125 - 30 y^{2} - 4 y^{3} = 0 \\ \Rightarrow 4 y^{3} + 30 y^{2} - 125 = 0 \\ ∴ Re q u i r e d E q u a t i o n i s 4 x^{3} + 30 x^{2} - 125 = 0 \end{array}$