α, β, γ are roots of x3−3x2+3x+7=0 ( ω is cube root of unity) then α−1β−1+β−1γ−1+γ−1α−1 is

x3−3x2+3x+7=0x3−13+31x−31x2+8=0x−13+8=0x−13=−8=−23x−1−23=13x−1−2=11/31,ω,ω2x−1=−2,−2ω.−2ω2x=−1,1−2ω,1−2ω2α−1β−1+β−1γ−1+γ−1α−1=−2−2ω+−2ω−2ω2+−2ω2−2=1ω+1ω+ω2=1ω+1ω+1ω=3ω

α, β, γ are roots of x3−3x2+3x+7=0 ( ω is cube root of unity) then α−1β−1+β−1γ−1+γ−1α−1 is

x3−3x2+3x+7=0x3−13+31x−31x2+8=0x−13+8=0x−13=−8=−23x−1−23=13x−1−2=11/31,ω,ω2x−1=−2,−2ω.−2ω2x=−1,1−2ω,1−2ω2α−1β−1+β−1γ−1+γ−1α−1=−2−2ω+−2ω−2ω2+−2ω2−2=1ω+1ω+ω2=1ω+1ω+1ω=3ω

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$α, β, γ$ are roots of $x^{3} - 3 x^{2} + 3 x + 7 = 0$ ( $ω$ is cube root of unity) then $(\frac{α - 1}{β - 1} + \frac{β - 1}{γ - 1} + \frac{γ - 1}{α - 1})$ is

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$\frac{3}{ω}$

$ω^{2}$

$2 ω^{2}$

$3 ω$

answer is A.

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Detailed Solution

$\begin{array}{l} x^{3} - 3 x^{2} + 3 x + 7 = 0 x^{3} - 1^{3} + 3 (1) (x) - 3 (1) (x^{2}) + 8 = 0 \\ {(x - 1)}^{3} + 8 = 0 {(x - 1)}^{3} = - 8 = - {(2)}^{3} \\ {(\frac{x - 1}{- 2})}^{3} = 1^{3} \frac{x - 1}{- 2} = 1^{1 / 3} (1, ω, ω^{2}) \\ x - 1 = - 2, - 2 ω . - 2 ω^{2} \\ x = - 1, 1 - 2 ω, 1 - 2 ω^{2} \\ \frac{α - 1}{β - 1} + \frac{β - 1}{γ - 1} + \frac{γ - 1}{α - 1} = \frac{- 2}{- 2 ω} + \frac{- 2 ω}{- 2 ω^{2}} + \frac{- 2 ω^{2}}{- 2} \\ = \frac{1}{ω} + \frac{1}{ω} + ω^{2} = \frac{1}{ω} + \frac{1}{ω} + \frac{1}{ω} = \frac{3}{ω} \end{array}$