If α,β,γ are the roots of x3–3x+1=0 then the equation whose roots are α−1βγ,β−1γα,γ−1αβ is

α−1βγ=α1−1αβγ=α1−1−1=2α⇒If α , β , γ are the roots of fx=x3–3x+1=0 then the equation whose roots areα−1βγ ,β−1γα,γ−1αβ (i.e)2α,2β,2γ are roots of fx2=0fx2=x23-3x2+1=0⇒x3-12x+8=0

If α,β,γ are the roots of x3–3x+1=0 then the equation whose roots are α−1βγ,β−1γα,γ−1αβ is

α−1βγ=α1−1αβγ=α1−1−1=2α⇒If α , β , γ are the roots of fx=x3–3x+1=0 then the equation whose roots areα−1βγ ,β−1γα,γ−1αβ (i.e)2α,2β,2γ are roots of fx2=0fx2=x23-3x2+1=0⇒x3-12x+8=0

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If $α, β, γ$ are the roots of x³–3x+1=0 then the equation whose roots are $α - \frac{1}{βγ}, β - \frac{1}{γα}, γ - \frac{1}{αβ}$ is

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

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

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

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

answer is D.

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Detailed Solution

$\begin{aligned} α - \frac{1}{βγ} = α (1 - \frac{1}{αβγ}) = α (1 - \frac{1}{- 1}) = 2 α \Rightarrow \end{aligned}$

$\begin{array}{r} I f α, β, γ a r e t h e r o o t s o f f (x) = x^{3} - 3 x + 1 = 0 \\ t h e n t h e e q u a t i o n w h o s e r o o t s a r e α - \frac{1}{βγ}, β - \frac{1}{γα}, γ - \frac{1}{αβ} (i . e) \\ 2 α, 2 β, 2 γ are roots of f (\frac{x}{2}) = 0 \\ f (\frac{x}{2}) = {(\frac{x}{2})}^{3} - 3 (\frac{x}{2}) + 1 = 0 \Rightarrow x^{3} - 12 x + 8 = 0 \end{array}$