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Question: If we have the roots as \(\alpha ,\beta ,\lambda \) of the equation, \({x^3} - 3{x^2} + 2x - 1 = 0\)...

If we have the roots as α,β,λ\alpha ,\beta ,\lambda of the equation, x3−3x2+2x−1=0{x^3} - 3{x^2} + 2x - 1 = 0, then the value of (1−α)(1−β)(1−λ)\left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \lambda } \right) is
(a)1\left( a \right)1
(b)2\left( b \right)2
(c)−1\left( c \right) - 1
(d)−2\left( d \right) - 2

Explanation

Solution

In this particular question use the concept that for a cubic equation the sum of the roots is the ratio of negative times the coefficient of x2{x^2} to the coefficient of x3{x^3} and the sum of the product of two roots is the ratio of the coefficient of x to the coefficient of x3{x^3}, and the product of the roots is the ratio of negative times the constant term to the coefficient of x3{x^3}, so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given cubic equation:
x3−3x2+2x−1=0{x^3} - 3{x^2} + 2x - 1 = 0
And α,β,λ\alpha,\beta,\lambda are the roots of the above equation.
Now consider a standard cubic equation ax3+bx2+cx+d=0a{x^3} + b{x^2} + cx + d = 0
Let p, q and r be the roots of the above standard equation.
Now as we know that the sum of the roots is the ratio of negative times the coefficient of x2{x^2} to the coefficient of x3{x^3}.
⇒p+q+r= - coefficient of x2coefficient of x3\Rightarrow p + q + r = \dfrac{{{\text{ - coefficient of }}{x^2}}}{{{\text{coefficient of }}{x^3}}}
⇒p+q+r=−ba\Rightarrow p + q + r = \dfrac{{ - b}}{a}
Now we also know that the sum of the product of two roots is the ratio of the coefficient of x to the coefficient of x3{x^3}.
⇒pq+qr+rp=coefficient of xcoefficient of x3\Rightarrow pq + qr + rp = \dfrac{{{\text{coefficient of }}x}}{{{\text{coefficient of }}{x^3}}}
⇒pq+qr+rp=ca\Rightarrow pq + qr + rp = \dfrac{c}{a}
Now we also know that the product of the roots is the ratio of negative times the constant term to the coefficient of x3{x^3}.
⇒pqr=-constant termcoefficient of x3\Rightarrow pqr = \dfrac{{{\text{-constant term}}}}{{{\text{coefficient of }}{x^3}}}
⇒pqr=-da\Rightarrow pqr = \dfrac{{\text{-d}}}{a}
Now compare the standard equation with the given equation we have,
⇒a=1,b=−3,c=2,d=−1\Rightarrow a = 1,b = - 3,c = 2,d = - 1
So the sum of the roots is,
⇒α+β+γ=−(−3)1=3\Rightarrow \alpha + \beta + \gamma = \dfrac{{ - \left( { - 3} \right)}}{1} = 3................ (1)
The sum of the product of two roots is
⇒αβ+βγ+γα=21=2\Rightarrow \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{2}{1} = 2................ (2)
And the product of the roots is
⇒αβγ=−(−1)1=1\Rightarrow \alpha \beta \gamma = \dfrac{{ - \left( { - 1} \right)}}{1} = 1................. (3)
Now we have to find out the value of
(1−α)(1−β)(1−λ)\left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \lambda } \right)
Now simplify this equation we have,
⇒(1−α)(1−β)(1−λ)=(1−α−β+αβ)(1−λ)\Rightarrow \left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \lambda } \right) = \left( {1 - \alpha - \beta + \alpha \beta } \right)\left( {1 - \lambda } \right)
⇒(1−α)(1−β)(1−λ)=(1−α−β+αβ−(λ−αλ−βλ+αβλ))\Rightarrow \left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \lambda } \right) = \left( {1 - \alpha - \beta + \alpha \beta - \left( {\lambda - \alpha \lambda - \beta \lambda + \alpha \beta \lambda } \right)} \right)
⇒(1−α)(1−β)(1−λ)=(1−α−β+αβ−λ+αλ+βλ−αβλ)\Rightarrow \left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \lambda } \right) = \left( {1 - \alpha - \beta + \alpha \beta - \lambda + \alpha \lambda + \beta \lambda - \alpha \beta \lambda } \right)
⇒(1−α)(1−β)(1−λ)=(1−(α+β+γ))+(αβ+αλ+βλ)−αβλ)\Rightarrow \left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \lambda } \right) = \left( {1 - \left( {\alpha + \beta + \gamma) } \right) + \left( {\alpha \beta + \alpha \lambda + \beta \lambda } \right) - \alpha \beta \lambda } \right)
Now substitute the values from equations (1), (2), and (3) in the above equation we have,
⇒(1−α)(1−β)(1−λ)=(1−(3)+(2)−1)\Rightarrow \left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \lambda } \right) = \left( {1 - \left( 3 \right) + \left( 2 \right) - 1} \right)
⇒(1−α)(1−β)(1−λ)=−1\Rightarrow \left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \lambda } \right) = - 1
So this is the required answer.
Hence option (c) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall how to crackdown the cubic equation in terms of his roots so first find out the sum of the roots, the sum of the product of the two roots and product of the roots of the cubic equation respectively as above then simplify the given equation whose value we have to find out, then substitute the values in the simplified equation as above we will get the required answer.