If - - and - are the roots of the equation x3-3 x2-3 x-7-0- and - is the cube root of unity- then the value of -1-1-1-y-1-Y-1-1 is equal to

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

α, β

and

γ

are the 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})

Q. If

α, β, γ

are roots of equation

x^{3} - x - 1 = 0

, then the equation whose roots are

\frac{1}{β + γ}, \frac{1}{γ + α}, \frac{1}{α + β}

is -

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If α,β and γ are the roots of the equation x3−3x2+3x+7=0, and ω is the cube root of unity, then the value of α−1β−1+β−1γ−1+γ−1α−1 is equal to

The correct option is A 3ω2x3−3x2+3x+7=0⇒(x−1)3=−8⇒x=−1, 1−2ω, 1−2ω2 Let α=−1, β=1−2ω, γ=1−2ω2 ⇒α−1β−1+β−1γ−1+γ−1α−1=3ω2

If $α, β$ and $γ$ are the roots of the equation $x^{3} - 3 x^{2} + 3 x + 7 = 0,$ and $ω$ is the cube root of unity, then the value of $\frac{α - 1}{β - 1} + \frac{β - 1}{γ - 1} + \frac{γ - 1}{α - 1}$ is equal to

The correct option is A $3 ω^{2}$
$x^{3} - 3 x^{2} + 3 x + 7 = 0 \Rightarrow (x - 1)^{3} = - 8 \Rightarrow x = - 1, 1 - 2 ω, 1 - 2 ω^{2}$
Let $α = - 1, β = 1 - 2 ω, γ = 1 - 2 ω^{2}$
$\Rightarrow \frac{α - 1}{β - 1} + \frac{β - 1}{γ - 1} + \frac{γ - 1}{α - 1} = 3 ω^{2}$