CAT 2024 Slot 2 QA Question 6: The roots $\alpha$, $\beta$ of the equation $3x^2 + \lambda x - 1 = 0$, satisfy $\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} = 15$. The value of $(\alpha^3 + \beta^3)^2$ is

Question

The roots $\alpha$, $\beta$ of the equation $3x^2 + \lambda x - 1 = 0$, satisfy $\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} = 15$. The value of $(\alpha^3 + \beta^3)^2$ is

Accepted Answer

For the equation $3x^2 + \lambda x - 1 = 0$, use sum and product of roots.
$\alpha + \beta = -\dfrac{\lambda}{3}$
$\alpha\beta = -\dfrac{1}{3}$
Using the given condition,
$\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} = \dfrac{\alpha^2+\beta^2}{\alpha^2\beta^2} = 15$
$\Rightarrow \dfrac{\alpha^2+\beta^2}

The roots α\alphaα, β\betaβ of the equation 3x2+λx−1=03x^2 + \lambda x - 1 = 03x2+λx−1=0, satisfy 1α2+1β2=15\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} = 15α21​+β21​=15. The value of (α3+β3)2(\alpha^3 + \beta^3)^2(α3+β3)2 is

The roots $\alpha$ , $\beta$ of the equation $3x^2 + \lambda x - 1 = 0$ , satisfy $\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} = 15$ . The value of $(\alpha^3 + \beta^3)^2$ is