If we have the roots as $\alpha ,\beta ,\lambda $ of the equation, ${x^3} - 3{x^2} + 2x - 1 = 0$, then the value of $\left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \lambda } \right)$ is
$\left( a \right)1$
$\left( b \right)2$
$\left( c \right) - 1$
$\left( d \right) - 2$
Answer
654.6k+ views
Hint: 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 ${x^2}$ to the coefficient of ${x^3}$ and the sum of the product of two roots is the ratio of the coefficient of x to the coefficient of ${x^3}$, and the product of the roots is the ratio of negative times the constant term to the coefficient of ${x^3}$, so use these concepts to reach the solution of the question.
Complete step-by-step solution:
Given cubic equation:
${x^3} - 3{x^2} + 2x - 1 = 0$
And $\alpha,\beta,\lambda $ are the roots of the above equation.
Now consider a standard cubic equation $a{x^3} + b{x^2} + cx + d = 0$
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 ${x^2}$ to the coefficient of ${x^3}$.
$ \Rightarrow p + q + r = \dfrac{{{\text{ - coefficient of }}{x^2}}}{{{\text{coefficient of }}{x^3}}}$
$ \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 ${x^3}$.
$ \Rightarrow pq + qr + rp = \dfrac{{{\text{coefficient of }}x}}{{{\text{coefficient of }}{x^3}}}$
$ \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 ${x^3}$.
$ \Rightarrow pqr = \dfrac{{{\text{-constant term}}}}{{{\text{coefficient of }}{x^3}}}$
$ \Rightarrow pqr = \dfrac{{\text{-d}}}{a}$
Now compare the standard equation with the given equation we have,
$ \Rightarrow a = 1,b = - 3,c = 2,d = - 1$
So the sum of the roots is,
$ \Rightarrow \alpha + \beta + \gamma = \dfrac{{ - \left( { - 3} \right)}}{1} = 3$................ (1)
The sum of the product of two roots is
$ \Rightarrow \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{2}{1} = 2$................ (2)
And the product of the roots is
$ \Rightarrow \alpha \beta \gamma = \dfrac{{ - \left( { - 1} \right)}}{1} = 1$................. (3)
Now we have to find out the value of
$\left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \lambda } \right)$
Now simplify this equation we have,
$ \Rightarrow \left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \lambda } \right) = \left( {1 - \alpha - \beta + \alpha \beta } \right)\left( {1 - \lambda } \right)$
$ \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)$
$ \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)$
\[ \Rightarrow \left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \lambda } \right) = \left( {1 - \left( {\alpha + \beta + \gamma) } \right) + \left( {\alpha \beta + \alpha \lambda + \beta \lambda } \right) - \alpha \beta \lambda } \right)\]
Now substitute the values from equations (1), (2), and (3) in the above equation we have,
\[ \Rightarrow \left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \lambda } \right) = \left( {1 - \left( 3 \right) + \left( 2 \right) - 1} \right)\]
\[ \Rightarrow \left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \lambda } \right) = - 1\]
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.
Complete step-by-step solution:
Given cubic equation:
${x^3} - 3{x^2} + 2x - 1 = 0$
And $\alpha,\beta,\lambda $ are the roots of the above equation.
Now consider a standard cubic equation $a{x^3} + b{x^2} + cx + d = 0$
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 ${x^2}$ to the coefficient of ${x^3}$.
$ \Rightarrow p + q + r = \dfrac{{{\text{ - coefficient of }}{x^2}}}{{{\text{coefficient of }}{x^3}}}$
$ \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 ${x^3}$.
$ \Rightarrow pq + qr + rp = \dfrac{{{\text{coefficient of }}x}}{{{\text{coefficient of }}{x^3}}}$
$ \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 ${x^3}$.
$ \Rightarrow pqr = \dfrac{{{\text{-constant term}}}}{{{\text{coefficient of }}{x^3}}}$
$ \Rightarrow pqr = \dfrac{{\text{-d}}}{a}$
Now compare the standard equation with the given equation we have,
$ \Rightarrow a = 1,b = - 3,c = 2,d = - 1$
So the sum of the roots is,
$ \Rightarrow \alpha + \beta + \gamma = \dfrac{{ - \left( { - 3} \right)}}{1} = 3$................ (1)
The sum of the product of two roots is
$ \Rightarrow \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{2}{1} = 2$................ (2)
And the product of the roots is
$ \Rightarrow \alpha \beta \gamma = \dfrac{{ - \left( { - 1} \right)}}{1} = 1$................. (3)
Now we have to find out the value of
$\left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \lambda } \right)$
Now simplify this equation we have,
$ \Rightarrow \left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \lambda } \right) = \left( {1 - \alpha - \beta + \alpha \beta } \right)\left( {1 - \lambda } \right)$
$ \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)$
$ \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)$
\[ \Rightarrow \left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \lambda } \right) = \left( {1 - \left( {\alpha + \beta + \gamma) } \right) + \left( {\alpha \beta + \alpha \lambda + \beta \lambda } \right) - \alpha \beta \lambda } \right)\]
Now substitute the values from equations (1), (2), and (3) in the above equation we have,
\[ \Rightarrow \left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \lambda } \right) = \left( {1 - \left( 3 \right) + \left( 2 \right) - 1} \right)\]
\[ \Rightarrow \left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \lambda } \right) = - 1\]
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.
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