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Question: If \(\alpha ,\beta ,\gamma \) are the roots of the equation \({x^3} + ax + b = 0\), then what is the...

If α,β,γ\alpha ,\beta ,\gamma are the roots of the equation x3+ax+b=0{x^3} + ax + b = 0, then what is the value of α3+β3+γ3α2+β2+γ2\dfrac{{{\alpha ^3} + {\beta ^3} + {\gamma ^3}}}{{{\alpha ^2} + {\beta ^2} + {\gamma ^2}}}.
A. 3b2a B. −3b2a C. 3b D. 2a  {\text{A}}{\text{. }}\dfrac{{3b}}{{2a}} \\\ {\text{B}}{\text{. }}\dfrac{{ - 3b}}{{2a}} \\\ {\text{C}}{\text{. }}3b \\\ {\text{D}}{\text{. }}2a \\\

Explanation

Solution

Hint- Here, we will proceed by using the formulas which are α+β+γ=−(Coefficient of x2)Coefficient of x3\alpha + \beta + \gamma = \dfrac{{ - \left( {{\text{Coefficient of }}{x^2}} \right)}}{{{\text{Coefficient of }}{x^3}}}, αβ+βγ+αγ=Coefficient of xCoefficient of x3\alpha \beta + \beta \gamma + \alpha \gamma = \dfrac{{{\text{Coefficient of }}x}}{{{\text{Coefficient of }}{x^3}}} and αβγ=−(Constant term)Coefficient of x3\alpha \beta \gamma = \dfrac{{ - \left( {{\text{Constant term}}} \right)}}{{{\text{Coefficient of }}{x^3}}} for any general cubic equation having three roots as α,β,γ\alpha ,\beta ,\gamma

“Complete step-by-step answer:”
Given cubic equation is x3+ax+b=0 →(1){x^3} + ax + b = 0{\text{ }} \to (1{\text{)}}
For any general cubic equation cx3+dx2+ex+f=0 →(2)c{x^3} + d{x^2} + ex + f = 0{\text{ }} \to {\text{(2)}} which have three roots as α,β,γ\alpha ,\beta ,\gamma ,
Sum of the roots, α+β+γ=−(Coefficient of x2)Coefficient of x3=−dc →(3)\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, αβ+βγ+αγ=Coefficient of xCoefficient of x3=ec →(4)\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, αβγ=−(Constant term)Coefficient of x3=−fc →(5)\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, α+β+γ=01=0\alpha + \beta + \gamma = \dfrac{0}{1} = 0
Sum of product of the roots taken two at a time, αβ+βγ+αγ=a1=a\alpha \beta + \beta \gamma + \alpha \gamma = \dfrac{a}{1} = a
Product of roots, αβγ=−b1=−b\alpha \beta \gamma = \dfrac{{ - b}}{1} = - b
As we know that x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−xz){x^3} + {y^3} + {z^3} - 3xyz = \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - xz} \right)
So, α3+β3+γ3−3αβγ=(α+β+γ)(α2+β2+γ2−αβ−βγ−αγ){\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 α+β+γ=0\alpha + \beta + \gamma = 0
⇒α3+β3+γ3−3αβγ=(0)(α2+β2+γ2−αβ−βγ−αγ) ⇒α3+β3+γ3−3αβγ=0 ⇒α3+β3+γ3=3αβγ  \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, αβγ=−b\alpha \beta \gamma = - b
⇒α3+β3+γ3=3(−b) ⇒α3+β3+γ3=−3b→(6)  \Rightarrow {\alpha ^3} + {\beta ^3} + {\gamma ^3} = 3\left( { - b} \right) \\\ \Rightarrow {\alpha ^3} + {\beta ^3} + {\gamma ^3} = - 3b \to {\text{(6)}} \\\
Also, (x+y+z)2=x2+y2+z2+2xy+2yz+2xz ⇒x2+y2+z2=(x+y+z)2−2(xy+yz+xz)  {\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) \\\
⇒α2+β2+γ2=(α+β+γ)2−2(αβ+βγ+αγ)\Rightarrow {\alpha ^2} + {\beta ^2} + {\gamma ^2} = {\left( {\alpha + \beta + \gamma } \right)^2} - 2\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right)
But α+β+γ=0\alpha + \beta + \gamma = 0 and αβ+βγ+αγ=a\alpha \beta + \beta \gamma + \alpha \gamma = a
⇒α2+β2+γ2=(0)2−2(a) ⇒α2+β2+γ2=−2a →(7)  \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
α3+β3+γ3α2+β2+γ2=−3b−2a=3b2a\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 α3+β3+γ3α2+β2+γ2\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 (α+β+γ)\left( {\alpha + \beta + \gamma } \right), (αβ+βγ+αγ)\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right) and αβγ\alpha \beta \gamma which can be easily obtained with the help of the known formulas for any general cubic equation.