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(A) Given the cubic equation $x^3 - ax^2 + bx - c = 0$ with roots $\alpha, \beta, \gamma$. By Vieta's formulas: $\alpha + \beta + \gamma = a$ $\alpha\

Vedclass · Accepted Answer

$\frac{b^2 - 2ac}{c^2}$

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(A) Given the cubic equation $x^3 - ax^2 + bx - c = 0$ with roots $\alpha, \beta, \gamma$. By Vieta's formulas: $\alpha + \beta + \gamma = a$ $\alpha\beta + \beta\gamma + \gamma\alpha = b$ $\alpha\beta\gamma = c$ We need to find $\alpha^{-2} + \beta^{-2} + \gamma^{-2} = \frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2}$. This can be written as: $\frac{\beta^2\gamma^2 + \alpha^2\gamma^2 + \alpha^2\beta^2}{(\alpha\beta\gamma)^2} = \frac{(\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma)}{(\alpha\beta\gamma)^2}$. Substituting the values: $= \frac{b^2 - 2(c)(a)}{c^2} = \frac{b^2 - 2ac}{c^2}$.

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3 - ax^2 + bx - c = 0$,then $\alpha^{-2} + \beta^{-2} + \gamma^{-2} = $

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