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Mathematics Question on Complex Numbers and Quadratic Equations

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

A

1

B

2

C

-1

D

-2

Answer

-1

Explanation

Solution

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