If α, β , γ are the roots of the equation x3 - 3x2 + 2x - 1 = 0 then the value of [(1 - α) (1 -β )(1 - γ)] is

Question

If α, β , &gamma; are the roots of the equation x3 - 3x2 + 2x - 1 = 0 then the value of [(1 - α) (1 -β )(1 - &gamma;)] is (A) 1 (B) 2 (C) -1 (D) -2

Tardigrade Admin · Accepted Answer

We have; α, β , &gamma; = 3, α β + β &gamma; +&gamma; α = 2 and α β &gamma; = 1 Now, ( 1 - α)(1 - β )(1- &gamma;) = (1 - β - α + α β ) (1 - &gamma;) = (1 - β - α + α β - &gamma; + β &gamma; + α &gamma; - α β &gamma;) = [1 - ( α + β + &gamma; ) + ( α β + β &gamma; + &gamma; α) - α β &gamma;] = [ 1 - 3 + 2 - 1] = - 1

Q. If $α, β, γ$ are the roots of the equation $x^{3} - 3 x^{2} + 2 x - 1 = 0$ then the value of $[(1 - α) (1 - β) (1 - γ)]$ is

1

2

-1

-2

We have; $α, β, γ = 3$ ,
$α β + β γ + γ α = 2$ and $α β γ = 1$
Now, $(1 - α) (1 - β) (1 - γ)$
$= (1 - β - α + α β) (1 - γ)$
$= (1 - β - α + α β - γ + β γ + α γ - α β γ)$
$= [1 - (α + β + γ) + (α β + β γ + γ α) - α β γ]$
$= [1 - 3 + 2 - 1] = - 1$

Q. If α,β,γ are the roots of the equation x3−3x2+2x−1=0 then the value of [(1−α)(1−β)(1−γ)] is

1

2

-1

-2

We have; α,β,γ=3, αβ+βγ+γα=2 and αβγ=1 Now, (1−α)(1−β)(1−γ) =(1−β−α+αβ)(1−γ) =(1−β−α+αβ−γ+βγ+αγ−αβγ) =[1−(α+β+γ)+(αβ+βγ+γα)−αβγ] =[1−3+2−1]=−1

Q. If $α, β, γ$ are the roots of the equation $x^{3} - 3 x^{2} + 2 x - 1 = 0$ then the value of $[(1 - α) (1 - β) (1 - γ)]$ is

We have; $α, β, γ = 3$ ,
$α β + β γ + γ α = 2$ and $α β γ = 1$
Now, $(1 - α) (1 - β) (1 - γ)$
$= (1 - β - α + α β) (1 - γ)$
$= (1 - β - α + α β - γ + β γ + α γ - α β γ)$
$= [1 - (α + β + γ) + (α β + β γ + γ α) - α β γ]$
$= [1 - 3 + 2 - 1] = - 1$