Question

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

• A. 18
• B. 24
• C. 36
• D. 96

Answer 1

Answer: (B)

Here $\alpha, \beta$ roots of equation
$3 x^{2}+\lambda x-1=0$
$\alpha+\beta=\frac{-\lambda}{3}, \alpha \beta=\frac{-1}{3} $
$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{\alpha^{2} \beta^{2}}=15 $
$\lambda^{2}=9$
Now $6\left(\alpha^{3}+\beta^{3}\right)^{2}=6\left((\alpha+\beta)\left((\alpha+\beta)^{2}-3 \alpha \beta\right)\right)^{2}$
$=6\left(\frac{\lambda^{2}}{9}\right)\left\{\frac{\lambda^{2}}{9}+1\right\}^{2}=24$