
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 \\
$
Answer
605.1k+ views
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.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 8 Maths: Engaging Questions & Answers for Success

Class 8 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

