Key Concepts and Formulas

• Vieta's Formulas: For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
• Algebraic Identity: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
• Algebraic Identity: $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha\beta)$

Step-by-Step Solution

Step 1: Apply Vieta's Formulas

The given quadratic equation is $3x^2 + \lambda x - 1 = 0$. We want to find expressions for the sum and product of the roots $\alpha$ and $\beta$ in terms of $\lambda$.

• Why this step? Vieta's formulas provide a direct link between the roots of the quadratic equation and its coefficients. This allows us to express the sum and product of the roots, which are fundamental building blocks for solving this problem, in terms of the unknown parameter $\lambda$.

Using Vieta's Formulas:

• Sum of roots: $\alpha + \beta = -\frac{\lambda}{3}$
• Product of roots: $\alpha\beta = -\frac{1}{3}$

Step 2: Use the given condition to find $\lambda^2$

We are given that $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 15$. We need to manipulate this equation to solve for $\lambda^2$.

• Why this step? The given condition provides a crucial relationship between the roots. By expressing this relationship in terms of $\alpha + \beta$ and $\alpha\beta$, we can substitute the expressions derived from Vieta's formulas and solve for the unknown $\lambda^2$.

Rewriting the given condition: $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{\alpha^2\beta^2} = 15$ Using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$: $\frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2} = 15$ Substituting the values of $\alpha + \beta$ and $\alpha\beta$ from Step 1: $\frac{\left(-\frac{\lambda}{3}\right)^2 - 2\left(-\frac{1}{3}\right)}{\left(-\frac{1}{3}\right)^2} = 15$ Simplifying: $\frac{\frac{\lambda^2}{9} + \frac{2}{3}}{\frac{1}{9}} = 15$ Multiplying both the numerator and denominator of the main fraction by 9: $\frac{\lambda^2 + 6}{1} = 15$ $\lambda^2 + 6 = 15$ $\lambda^2 = 9$

Step 3: Calculate $\alpha^3 + \beta^3$

We need to find an expression for $\alpha^3 + \beta^3$ in terms of $\lambda$.

• Why this step? We need to evaluate $6(\alpha^3 + \beta^3)^2$, so finding an expression for $\alpha^3 + \beta^3$ in terms of $\lambda$ is a necessary intermediate step.

Using the identity $\alpha^3 + \beta^3 = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha\beta)$: $\alpha^3 + \beta^3 = \left(-\frac{\lambda}{3}\right)\left(\left(-\frac{\lambda}{3}\right)^2 - 3\left(-\frac{1}{3}\right)\right)$ $\alpha^3 + \beta^3 = \left(-\frac{\lambda}{3}\right)\left(\frac{\lambda^2}{9} + 1\right)$

Step 4: Calculate $6(\alpha^3 + \beta^3)^2$

Now we calculate the final expression.

• Why this step? This is the final calculation to arrive at the answer. Substituting the previously calculated value of $\lambda^2$ allows us to obtain a numerical result.

Substituting $\lambda^2 = 9$ into the expression for $\alpha^3 + \beta^3$: $\alpha^3 + \beta^3 = \left(-\frac{\lambda}{3}\right)\left(\frac{9}{9} + 1\right) = \left(-\frac{\lambda}{3}\right)(2) = -\frac{2\lambda}{3}$ Now, we calculate $6(\alpha^3 + \beta^3)^2$: $6(\alpha^3 + \beta^3)^2 = 6\left(-\frac{2\lambda}{3}\right)^2 = 6\left(\frac{4\lambda^2}{9}\right)$ Substituting $\lambda^2 = 9$: $6\left(\frac{4 \cdot 9}{9}\right) = 6(4) = 24$

Common Mistakes & Tips

• Be very careful with signs when applying Vieta's formulas and algebraic identities.
• Remember the correct formulas for the sum and product of roots.
• Double-check your algebraic manipulations to avoid errors.

Summary

We used Vieta's formulas to express the sum and product of the roots in terms of $\lambda$. Then, we used the given condition to find the value of $\lambda^2$. Finally, we substituted these values into the expression $6(\alpha^3 + \beta^3)^2$ to find its value, which is 24.

The final answer is \boxed{24}, which corresponds to option (B).