JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Quadratic Equations and Inequalities

For $3x^2+\lambda x-1=0$ the roots satisfy $\alpha +\beta =-\frac{\lambda }{3}$ and $\alpha \beta =-\frac{1}{3}$. The condition on the reciprocals reads $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=\frac{(\alpha +\beta {)}^2-2\alpha \beta }{(\alpha \beta {)}^2}$, and substituting the symmetric functions gives $\frac{\frac{{\lambda }^2}{9}+\frac{2}{3}}{\frac{1}{9}}={\lambda }^2+6$. Setting this equal to $15$ forces ${\lambda }^2=9$. Now ${\alpha }^3+{\beta }^3=(\alpha +\beta )[(\alpha +\beta {)}^2-3\alpha \beta ]=-\frac{\lambda }{3}(\frac{{\lambda }^2}{9}+1)$, and with ${\lambda }^2=9$ the bracket equals $1+1=2$, so ${\alpha }^3+{\beta }^3=-\frac{2\lambda }{3}$. Squaring removes the sign to give $({\alpha }^3+{\beta }^3{)}^2=\frac{4{\lambda }^2}{9}=\frac{4\cdot 9}{9}=4$, and therefore $6({\alpha }^3+{\beta }^3{)}^2=24$.