Question

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

• A. 16
• B. 9
• C. 1
• D. 4

Answer 1

Answer: (D)

4

Answer 2

From the given equation, we have: $\alpha + \beta = -\frac{\lambda}{3}$ $\alpha\beta = -\frac{1}{3}$

Now, we can use the given condition: $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 15 \implies \frac{\alpha^2 + \beta^2}{\alpha^2\beta^2} = 15$

Substituting $\alpha\beta = -\frac{1}{3}$, we get: $\alpha^2 + \beta^2 = -5$

We know that: $(\alpha + \beta)^3 = \alpha^3 + \beta^3 + 3\alpha\beta(\alpha + \beta)$

Substituting the values of $\alpha + \beta$ and $\alpha\beta$, we get: $(-\frac{\lambda}{3})^3 = \alpha^3 + \beta^3 + 3(-\frac{1}{3})(-\frac{\lambda}{3})$

Simplifying, we get: $\alpha^3 + \beta^3 = -\frac{\lambda^3}{27} + \frac{\lambda}{9}$

Now, we need to find the value of λ. We can use the identity: $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$

Substituting the values, we get: $(-\frac{\lambda}{3})^2 = -5 + 2(-\frac{1}{3})$

Solving for λ, we get $\lambda = \pm 3\sqrt{2}$.

Substituting the value of λ in the expression for $\alpha^3 + \beta^3$, we get: $(\alpha^3 + \beta^3)^2 = 4$

Therefore, the value of $(\alpha^3 + \beta^3)^2$ is 4.