Question

If $\alpha$ , $\beta$ and $\gamma$ are the roots of the equation ${{x}^{3}}-6{{x}^{2}}+11x+6=0$, then $\sum{{{\alpha }^{2}}\beta +}\sum{\alpha {{\beta }^{2}}}$ is equal to

1. $80$
2. $168$
3. $90$
4. $84$

Answer 1

In this type of question you need to use the given equations to find the sum of roots of the given equation, then sum of the product of roots of the equation and the product of roots of the given equation then we will use a simple algebraic formula to calculate the required value.

Complete step by step answer:
We have given the equation:
${{x}^{3}}-6{{x}^{2}}+11x+6=0$
Since it is a cubic equation therefore it can be of the form:
$A{{x}^{3}}+B{{x}^{2}}+Cx+D=0$
Where,
$A=1,B=-6,C=11,D=6$
Now it is given that the roots of the above equation are $\alpha$ , $\beta$ and $\gamma$ .
Now here we need to apply the formula of sum of roots, sum of product of roots and product of roots in order to find the value asked in the question.
So before applying it directly first we must know what it is
An equation in which at least one term is raised to the power of cubic but no term is raised to any higher power is called a cubic equation. The general form of the cubic equation is $A{{x}^{3}}+B{{x}^{2}}+Cx+D=0$.
So, the sum of roots of the cubic equation is given as:
$\alpha +\beta +\gamma =\dfrac{-B}{A}$
Where, $B$ is the coefficient of ${{x}^{2}}$ and $A$ is the coefficient of ${{x}^{3}}$ .
So now, the sum of product of roots of the cubic equation is given as:
$\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{C}{A}$
Where, $C$ is the coefficient of $x$ and $A$ is the coefficient of ${{x}^{3}}$ .
And the product of roots of the cubic equation is given as:
$\alpha \beta \gamma =\dfrac{-D}{A}$
Where, $D$ is the constant term and $A$ is the coefficient of ${{x}^{3}}$ .
Now we can apply the above formulas to find the required quantities and it may help in finding the value asked in the question.
So the sum of roots:
$\alpha +\beta +\gamma =\dfrac{-B}{A}$
Since we have, $A=1,B=-6,C=11,D=6$
$\Rightarrow \alpha +\beta +\gamma =6$
So the sum of product of roots:
$\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{C}{A}$
Since we have, $A=1,B=-6,C=11,D=6$
$\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =11$
And finally the product of roots:
$\alpha \beta \gamma =\dfrac{-D}{A}$
Since we have, $A=1,B=-6,C=11,D=6$
$\alpha \beta \gamma =-6$
$\Rightarrow \alpha +\beta +\gamma =6$
Now according to the question we need to find the value of:
$\sum{{{\alpha }^{2}}\beta +}\sum{\alpha {{\beta }^{2}}}$
So to find the above value we must use the formula:
$\sum{{{x}^{2}}y+}\sum{x{{y}^{2}}=}(x+y+z)(xy+yz+zx)-3(xyz)$
So using above formula we get,
$\Rightarrow \sum{{{\alpha }^{2}}\beta +}\sum{\alpha {{\beta }^{2}}=}(\alpha +\beta +\gamma )(\alpha \beta +\beta \gamma +\gamma \alpha )-3(\alpha \beta \gamma )$
Putting values of the quantities we found we get,
$\Rightarrow \sum{{{\alpha }^{2}}\beta +}\sum{\alpha {{\beta }^{2}}=}(6)(11)-3(-6)$
On solving further,

& \Rightarrow \sum{{{\alpha }^{2}}\beta +}\sum{\alpha {{\beta }^{2}}=}66+18 \\\ & \Rightarrow \sum{{{\alpha }^{2}}\beta +}\sum{\alpha {{\beta }^{2}}=}84 \\\ \end{aligned}$$ Hence we got the value of $$\sum{{{\alpha }^{2}}\beta +}\sum{\alpha {{\beta }^{2}}=}84$$ **Therefore our final answer is option $$(4)$$.** **Note:** We use cube and cube root in daily life. Example: Cube roots are used in day to day mathematics like in powers and exponents or to find the side of a three dimensional cube when its volume is given. Although it plays a great role in algebraic identities also.