Question

If we have the roots $\alpha ,\beta $ and $\gamma $ of the equation ${{x}^{3}}-8x+8=0$, then the values of $\sum{{{\alpha }^{2}}}$ and $\sum{\dfrac{1}{\alpha \beta }}$ are respectively
A. 0 and -16
B. 16 and 8
C. -16 and 0
D. 16 and 0

Answer 1

Hint: We find the relation between the roots and the coefficients of the equation using the formulas $\sum{x}=m+n+p=\dfrac{-b}{a}$, $\sum{xy=mn+np+mp=\dfrac{c}{a}}$, $\sum{xyz=mnp=\dfrac{-d}{a}}$ for general equation $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$. We use quadratic identities of \[{{\left( \alpha +\beta +\gamma \right)}^{2}}\] to form the equations with $\sum{{{\alpha }^{2}}}$ and $\sum{\dfrac{1}{\alpha \beta }}$. We place the values of the coordinates to find the solution of the problem.

Complete step-by-step solution
It’s given that $\alpha ,\beta ,\gamma $ are the roots of the equation ${{x}^{3}}-8x+8=0$.
The cubic equation has 3 roots and we know the relation between the roots and the coefficients of the equation.
For general equation form of cubic equation $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$ if we get three roots $m,n,p$ then we can say that the relations are
$\sum{x}=m+n+p=\dfrac{-b}{a}$, $\sum{xy=mn+np+mp=\dfrac{c}{a}}$, $\sum{xyz=mnp=\dfrac{-d}{a}}$.
We equate the given equation with the general equation to get $a=1,b=0,c=-8,d=8$.
Now we try to find the relations
$\sum{x}=\alpha +\beta +\gamma =0$, $\sum{xy=\alpha \beta +\beta \gamma +\alpha \gamma =-8}$, $\sum{xyz=\alpha \beta \gamma =-8}$.
We need to find the values of $\sum{{{\alpha }^{2}}}$ and $\sum{\dfrac{1}{\alpha \beta }}$.
We have the identities \[{{\left( \sum{x} \right)}^{2}}=\sum{{{x}^{2}}}+2\sum{xy}\].
So, ${{\left( \alpha +\beta +\gamma \right)}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}+2\alpha \beta +2\beta \gamma +2\alpha \gamma $.
Putting the values, we get
\[\begin{align}
  & {{\left( \alpha +\beta +\gamma \right)}^{2}}=\left( {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}} \right)+2\left( \alpha \beta +\beta \gamma +\alpha \gamma \right) \\
 & \Rightarrow 0=\sum{{{\alpha }^{2}}}+2\left( -8 \right) \\
 & \Rightarrow \sum{{{\alpha }^{2}}}=16 \\
\end{align}\]
Now, $\sum{\dfrac{1}{\alpha \beta }}=\dfrac{1}{\alpha \beta }+\dfrac{1}{\beta \gamma }+\dfrac{1}{\alpha \gamma }=\dfrac{\alpha +\beta +\gamma }{\alpha \beta \gamma }$
We put the values again to find $\sum{\dfrac{1}{\alpha \beta }}=\dfrac{\alpha +\beta +\gamma }{\alpha \beta \gamma }=\dfrac{0}{-8}=0$.
So, the values of $\sum{{{\alpha }^{2}}}$ and $\sum{\dfrac{1}{\alpha \beta }}$ are respectively 16 and 0. Hence The correct option is D.

Note: The way, the summations are written is just the notion of expression. $\sum{{{\alpha }^{2}}}$ and $\sum{\dfrac{1}{\alpha \beta }}$ expresses the sum of squares and the inverse of multiple of two roots taking one at a time. The expression of $\sum{xy}$ represents the sum of the multiplication of roots taken two at a time. Even though we have 3 roots and we are representing only 2, it represents the way the summation has been taken. Same goes for $\sum{x}$. This expresses the sum of the roots.