Answer
Verified
416.4k+ 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
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Which type of bond is stronger ionic or covalent class 12 chemistry CBSE
What organs are located on the left side of your body class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
When people say No pun intended what does that mea class 8 english CBSE