CAT 2024 Slot 2 · Quant · Question 16

From the coefficients, \alpha +\beta =-\frac{\lambda}{3} and \alpha \beta =-\frac{1}{3}.

Combine the given condition over a common denominator: \frac{1}{{\alpha}^{2}}+\frac{1}{{\beta}^{2}}=\frac{{\alpha}^{2}+{\beta}^{2}}{{\left( \alpha \beta \right)}^{2}}.

Now {\alpha}^{2}+{\beta}^{2}={\left( \alpha +\beta \right)}^{2}-2\alpha \beta =\frac{{\lambda}^{2}}{9}+\frac{2}{3}.

Also {\left( \alpha \beta \right)}^{2}=\frac{1}{9}.

So the condition reads \frac{\frac{{\lambda}^{2}}{9}+\frac{2}{3}}{\frac{1}{9}}={\lambda}^{2}+6=15.

Therefore {\lambda}^{2}=9.

For the required expression, {\alpha}^{3}+{\beta}^{3}={\left( \alpha +\beta \right)}^{3}-3\alpha \beta \left( \alpha +\beta \right).

Substituting, {\alpha}^{3}+{\beta}^{3}=-\frac{{\lambda}^{3}}{27}-3\times \left( -\frac{1}{3}\right)\times \left( -\frac{\lambda}{3}\right)=-\frac{{\lambda}^{3}}{27}-\frac{\lambda}{3}.

With \lambda =3 this is -1-1=-2, and with \lambda =-3 it is 1+1=2.

Either way the square is {\left( \pm 2 \right)}^{2}=4, which is the third option.

Shortcut. Since only {\lambda}^{2} is pinned down, the sign of \lambda stays unknown — but the question asks for a square, so the ambiguity cannot matter. Pick \lambda =3 and stop.