Question 6.

From the sum and product of roots, we get: $\alpha\ +\beta\ =-\dfrac{\lambda}{3}$ and $\alpha\ \beta\ =-\dfrac{1}{3}$

Simplifying the expression given in the question, we get: $\dfrac{\alpha^2+\beta^2\ }{\alpha^2\beta^2\ }=15$

and substituting the denominator's value as 1/9, we get:$\alpha^2+\beta^2\ =\dfrac{15}{9}$

We want the expression $\alpha^3+\beta^3\$, so multiplying both sides by $\alpha+\beta$, we get:

$\alpha^3+\beta^3+\alpha\beta\left(a+\beta\ \right)=\dfrac{15}{9}\left(\alpha\ +\beta\ \right)$

$\alpha^3+\beta^3+\dfrac{\lambda}{9}\ =\dfrac{15}{9}\left(-\dfrac{\lambda}{3}\ \right)$

$\alpha^3+\beta^3+\dfrac{\lambda}{9}\ =-\dfrac{5\lambda}{9}-\dfrac{\lambda}{9}=-\dfrac{2\lambda\ }{3}\ \$

We would still need to find the value of $\lambda$

This we can do from the initial relation we had:

$\alpha^2+\beta^2\ =\dfrac{15}{9}$

$\alpha^2+\beta^2\ =\left(\alpha+\beta\right)^2-2\alpha\ \beta\ \ \ =\dfrac{15}{9}$

$\dfrac{\lambda^2}{9}+\frac{2}{3}\ \ \ =\dfrac{15}{9}$

$\dfrac{\lambda^2}{9}\ =\dfrac{15-6}{9}=\dfrac{9}{9}=1$

This would finally give us $\lambda^2=9$

Using this in our required expression, we get:

$\left(\alpha^3+\beta^3\right)^2=\left(-\dfrac{2\lambda}{3}\ \ \right)^2=\dfrac{4\times\ 9}{9}=4$

Therefore, Option C is the correct answer.