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

We have a quadratic equation $3x^2 + \lambda x - 1 = 0$ with roots $\alpha$ and $\beta$. We're told that $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 15$, and we need to find $(\alpha^3 + \beta^3)^2$.

The key insight is to use Vieta's formulas - these connect the coefficients of a quadratic to its roots without actually solving for the roots.

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For any quadratic $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$:

Sum of roots: $\alpha + \beta = -\frac{b}{a}$

Product of roots: $\alpha\beta = \frac{c}{a}$

For our equation $3x^2 + \lambda x - 1 = 0$:

$\alpha + \beta = -\frac{\lambda}{3}$

$\alpha\beta = \frac{-1}{3} = -\frac{1}{3}$

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We're given: $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 15$

Getting a common denominator:

$\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{\alpha^2\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2}$

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We need to find $\alpha^2 + \beta^2$. Using the identity:

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$

Substituting our values:

$\alpha^2 + \beta^2 = \left(-\frac{\lambda}{3}\right)^2 - 2\left(-\frac{1}{3}\right)

= \frac{\lambda^2}{9} + \frac{2}{3}$

Also: $(\alpha\beta)^2 = \left(-\frac{1}{3}\right)^2 = \frac{1}{9}$

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Substituting into our condition:

$\frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = 15$

$\frac{\frac{\lambda^2}{9} + \frac{2}{3}}{\frac{1}{9}} = 15$

$\left(\frac{\lambda^2}{9} + \frac{2}{3}\right) \times 9 = 15$

$\lambda^2 + 6 = 15$

$\lambda^2 = 9$

Therefore: $\lambda = \pm 3$

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We use the identity: $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$

Case 1: $\lambda = 3$

$\alpha + \beta = -1$, $\alpha\beta = -\frac{1}{3}$

$\alpha^3 + \beta^3 = (-1)^3 - 3\left(-\frac{1}{3}\right)(-1) = -1 - 1 = -2$

Case 2: $\lambda = -3$

$\alpha + \beta = 1$, $\alpha\beta = -\frac{1}{3}$

$\alpha^3 + \beta^3 = (1)^3 - 3\left(-\frac{1}{3}\right)(1) = 1 + 1 = 2$

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For both cases:

When $\lambda = 3$: $(\alpha^3 + \beta^3)^2 = (-2)^2 = 4$

When $\lambda = -3$: $(\alpha^3 + \beta^3)^2 = (2)^2 = 4$

Both values of $\lambda$ give us the same final answer because we're squaring the result, which eliminates the sign difference.

Therefore, $(\alpha^3 + \beta^3)^2 = 4$