# If $\alpha ,\beta ,\gamma $ are the roots of the equation ${x^3} + ax + b = 0$, then what is the value of $\dfrac{{{\alpha ^3} + {\beta ^3} + {\gamma ^3}}}{{{\alpha ^2} + {\beta ^2} + {\gamma ^2}}}$.

$

{\text{A}}{\text{. }}\dfrac{{3b}}{{2a}} \\

{\text{B}}{\text{. }}\dfrac{{ - 3b}}{{2a}} \\

{\text{C}}{\text{. }}3b \\

{\text{D}}{\text{. }}2a \\

$

Last updated date: 19th Mar 2023

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Hint- Here, we will proceed by using the formulas which are $\alpha + \beta + \gamma = \dfrac{{ - \left( {{\text{Coefficient of }}{x^2}} \right)}}{{{\text{Coefficient of }}{x^3}}}$, $\alpha \beta + \beta \gamma + \alpha \gamma = \dfrac{{{\text{Coefficient of }}x}}{{{\text{Coefficient of }}{x^3}}}$ and $\alpha \beta \gamma = \dfrac{{ - \left( {{\text{Constant term}}} \right)}}{{{\text{Coefficient of }}{x^3}}}$ for any general cubic equation having three roots as $\alpha ,\beta ,\gamma $

“Complete step-by-step answer:”

Given cubic equation is ${x^3} + ax + b = 0{\text{ }} \to (1{\text{)}}$

For any general cubic equation $c{x^3} + d{x^2} + ex + f = 0{\text{ }} \to {\text{(2)}}$ which have three roots as $\alpha ,\beta ,\gamma $,

Sum of the roots, $\alpha + \beta + \gamma = \dfrac{{ - \left( {{\text{Coefficient of }}{x^2}} \right)}}{{{\text{Coefficient of }}{x^3}}} = \dfrac{{ - d}}{c}{\text{ }} \to {\text{(3)}}$

Sum of product of the roots taken two at a time, $\alpha \beta + \beta \gamma + \alpha \gamma = \dfrac{{{\text{Coefficient of }}x}}{{{\text{Coefficient of }}{x^3}}} = \dfrac{e}{c}{\text{ }} \to {\text{(4)}}$

Product of roots, $\alpha \beta \gamma = \dfrac{{ - \left( {{\text{Constant term}}} \right)}}{{{\text{Coefficient of }}{x^3}}} = \dfrac{{ - f}}{c}{\text{ }} \to {\text{(5)}}$

By comparing the given cubic equation (i.e., equation (1)) with the general cubic equation (i.e., equation (2)), we get

c=1, d=0, e=a and f=b

Putting the above obtained values, equations (3), (4) and (5) becomes

Sum of the roots, $\alpha + \beta + \gamma = \dfrac{0}{1} = 0$

Sum of product of the roots taken two at a time, $\alpha \beta + \beta \gamma + \alpha \gamma = \dfrac{a}{1} = a$

Product of roots, \[\alpha \beta \gamma = \dfrac{{ - b}}{1} = - b\]

As we know that ${x^3} + {y^3} + {z^3} - 3xyz = \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - xz} \right)$

So, ${\alpha ^3} + {\beta ^3} + {\gamma ^3} - 3\alpha \beta \gamma = \left( {\alpha + \beta + \gamma } \right)\left( {{\alpha ^2} + {\beta ^2} + {\gamma ^2} - \alpha \beta - \beta \gamma - \alpha \gamma } \right)$

But $\alpha + \beta + \gamma = 0$

$

\Rightarrow {\alpha ^3} + {\beta ^3} + {\gamma ^3} - 3\alpha \beta \gamma = \left( 0 \right)\left( {{\alpha ^2} + {\beta ^2} + {\gamma ^2} - \alpha \beta - \beta \gamma - \alpha \gamma } \right) \\

\Rightarrow {\alpha ^3} + {\beta ^3} + {\gamma ^3} - 3\alpha \beta \gamma = 0 \\

\Rightarrow {\alpha ^3} + {\beta ^3} + {\gamma ^3} = 3\alpha \beta \gamma \\

$

As, \[\alpha \beta \gamma = - b\]

$

\Rightarrow {\alpha ^3} + {\beta ^3} + {\gamma ^3} = 3\left( { - b} \right) \\

\Rightarrow {\alpha ^3} + {\beta ^3} + {\gamma ^3} = - 3b \to {\text{(6)}} \\

$

Also, $

{\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2xy + 2yz + 2xz \\

\Rightarrow {x^2} + {y^2} + {z^2} = {\left( {x + y + z} \right)^2} - 2\left( {xy + yz + xz} \right) \\

$

$ \Rightarrow {\alpha ^2} + {\beta ^2} + {\gamma ^2} = {\left( {\alpha + \beta + \gamma } \right)^2} - 2\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right)$

But $\alpha + \beta + \gamma = 0$ and $\alpha \beta + \beta \gamma + \alpha \gamma = a$

$

\Rightarrow {\alpha ^2} + {\beta ^2} + {\gamma ^2} = {\left( 0 \right)^2} - 2\left( a \right) \\

\Rightarrow {\alpha ^2} + {\beta ^2} + {\gamma ^2} = - 2a{\text{ }} \to {\text{(7)}} \\

$

Using equations (6) and (7), we get

$\dfrac{{{\alpha ^3} + {\beta ^3} + {\gamma ^3}}}{{{\alpha ^2} + {\beta ^2} + {\gamma ^2}}} = \dfrac{{ - 3b}}{{ - 2a}} = \dfrac{{3b}}{{2a}}$

Hence, option A is correct.

Note- In this particular problem, we have converted the expression $\dfrac{{{\alpha ^3} + {\beta ^3} + {\gamma ^3}}}{{{\alpha ^2} + {\beta ^2} + {\gamma ^2}}}$ whose value is required in terms of the known values which are $\left( {\alpha + \beta + \gamma } \right)$, $\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right)$ and \[\alpha \beta \gamma \] which can be easily obtained with the help of the known formulas for any general cubic equation.

“Complete step-by-step answer:”

Given cubic equation is ${x^3} + ax + b = 0{\text{ }} \to (1{\text{)}}$

For any general cubic equation $c{x^3} + d{x^2} + ex + f = 0{\text{ }} \to {\text{(2)}}$ which have three roots as $\alpha ,\beta ,\gamma $,

Sum of the roots, $\alpha + \beta + \gamma = \dfrac{{ - \left( {{\text{Coefficient of }}{x^2}} \right)}}{{{\text{Coefficient of }}{x^3}}} = \dfrac{{ - d}}{c}{\text{ }} \to {\text{(3)}}$

Sum of product of the roots taken two at a time, $\alpha \beta + \beta \gamma + \alpha \gamma = \dfrac{{{\text{Coefficient of }}x}}{{{\text{Coefficient of }}{x^3}}} = \dfrac{e}{c}{\text{ }} \to {\text{(4)}}$

Product of roots, $\alpha \beta \gamma = \dfrac{{ - \left( {{\text{Constant term}}} \right)}}{{{\text{Coefficient of }}{x^3}}} = \dfrac{{ - f}}{c}{\text{ }} \to {\text{(5)}}$

By comparing the given cubic equation (i.e., equation (1)) with the general cubic equation (i.e., equation (2)), we get

c=1, d=0, e=a and f=b

Putting the above obtained values, equations (3), (4) and (5) becomes

Sum of the roots, $\alpha + \beta + \gamma = \dfrac{0}{1} = 0$

Sum of product of the roots taken two at a time, $\alpha \beta + \beta \gamma + \alpha \gamma = \dfrac{a}{1} = a$

Product of roots, \[\alpha \beta \gamma = \dfrac{{ - b}}{1} = - b\]

As we know that ${x^3} + {y^3} + {z^3} - 3xyz = \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - xz} \right)$

So, ${\alpha ^3} + {\beta ^3} + {\gamma ^3} - 3\alpha \beta \gamma = \left( {\alpha + \beta + \gamma } \right)\left( {{\alpha ^2} + {\beta ^2} + {\gamma ^2} - \alpha \beta - \beta \gamma - \alpha \gamma } \right)$

But $\alpha + \beta + \gamma = 0$

$

\Rightarrow {\alpha ^3} + {\beta ^3} + {\gamma ^3} - 3\alpha \beta \gamma = \left( 0 \right)\left( {{\alpha ^2} + {\beta ^2} + {\gamma ^2} - \alpha \beta - \beta \gamma - \alpha \gamma } \right) \\

\Rightarrow {\alpha ^3} + {\beta ^3} + {\gamma ^3} - 3\alpha \beta \gamma = 0 \\

\Rightarrow {\alpha ^3} + {\beta ^3} + {\gamma ^3} = 3\alpha \beta \gamma \\

$

As, \[\alpha \beta \gamma = - b\]

$

\Rightarrow {\alpha ^3} + {\beta ^3} + {\gamma ^3} = 3\left( { - b} \right) \\

\Rightarrow {\alpha ^3} + {\beta ^3} + {\gamma ^3} = - 3b \to {\text{(6)}} \\

$

Also, $

{\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2xy + 2yz + 2xz \\

\Rightarrow {x^2} + {y^2} + {z^2} = {\left( {x + y + z} \right)^2} - 2\left( {xy + yz + xz} \right) \\

$

$ \Rightarrow {\alpha ^2} + {\beta ^2} + {\gamma ^2} = {\left( {\alpha + \beta + \gamma } \right)^2} - 2\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right)$

But $\alpha + \beta + \gamma = 0$ and $\alpha \beta + \beta \gamma + \alpha \gamma = a$

$

\Rightarrow {\alpha ^2} + {\beta ^2} + {\gamma ^2} = {\left( 0 \right)^2} - 2\left( a \right) \\

\Rightarrow {\alpha ^2} + {\beta ^2} + {\gamma ^2} = - 2a{\text{ }} \to {\text{(7)}} \\

$

Using equations (6) and (7), we get

$\dfrac{{{\alpha ^3} + {\beta ^3} + {\gamma ^3}}}{{{\alpha ^2} + {\beta ^2} + {\gamma ^2}}} = \dfrac{{ - 3b}}{{ - 2a}} = \dfrac{{3b}}{{2a}}$

Hence, option A is correct.

Note- In this particular problem, we have converted the expression $\dfrac{{{\alpha ^3} + {\beta ^3} + {\gamma ^3}}}{{{\alpha ^2} + {\beta ^2} + {\gamma ^2}}}$ whose value is required in terms of the known values which are $\left( {\alpha + \beta + \gamma } \right)$, $\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right)$ and \[\alpha \beta \gamma \] which can be easily obtained with the help of the known formulas for any general cubic equation.

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