Question

If the sum of the squares of the reciprocals of the roots $\alpha$ and $\beta$ of the equation $3x^2+\lambda x-1=0$ is $15,$ then $6{({\alpha }^3+{\beta }^3)}^2$ is equal to

Answer 1

Given, $\alpha$ and $\beta$ are roots of the equation $3x^2+\lambda x-1=0.$
$\therefore \alpha +\beta =-\frac{\lambda }{3}$ and $\alpha \beta =-\frac{1}{3}$
Also, $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=15$ $\Rightarrow \frac{(\alpha +\beta {)}^2-2\alpha \beta }{{\alpha }^2{\beta }^2}=15$
$\Rightarrow \frac{\frac{{\lambda }^2}{9}+\frac{2}{3}}{\frac{1}{9}}=15$
$\Rightarrow {\lambda }^2=9$
Now, $({\alpha }^3+{\beta }^3{)}^2$
$={\left[\right(\alpha +\beta \left)\left(\right(\alpha +\beta {)}^2-3\alpha \beta )\right]}^2$
$={\left[(-\frac{\lambda }{3})(\frac{{\lambda }^2}{9}-3(-\frac{1}{3}))\right]}^2$
$=\frac{{\lambda }^2}{9}{(\frac{{\lambda }^2}{9}+1)}^2$
$=4$
Hence, the value of $6{({\alpha }^3+{\beta }^3)}^2$ is $24.$