If alpha, beta, gamma and delta are zeroes of | vedclass

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(B) Given the polynomial equation: $x^4-3x^2+6x-12=0$ ... $(i)$ Comparing this with $ax^4+bx^3+cx^2+dx+e=0$,we get $a=1, b=0, c=-3, d=6, e=-12$. Since

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$\frac{-1}{2}$

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(B) Given the polynomial equation: $x^4-3x^2+6x-12=0$ ... $(i)$ Comparing this with $ax^4+bx^3+cx^2+dx+e=0$,we get $a=1, b=0, c=-3, d=6, e=-12$. Since $\alpha, \beta, \gamma, \delta$ are the roots,by Vieta's formulas: $\alpha+\beta+\gamma+\delta = -\frac{b}{a} = 0$. This implies $\alpha+\beta+\gamma = -\delta$,$\alpha+\delta+\gamma = -\beta$,$\alpha+\beta+\delta = -\gamma$,and $\delta+\beta+\gamma = -\alpha$. Also,$\Sigma \alpha\beta\gamma = -\frac{d}{a} = -6$ and $\alpha\beta\gamma\delta = \frac{e}{a} = -12$. Substituting these into the given expression: $\frac{-\delta}{\delta^2} + \frac{-\beta}{\beta^2} + \frac{-\gamma}{\gamma^2} + \frac{-\alpha}{\alpha^2} = -(\frac{1}{\delta} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\alpha})$. This simplifies to: $-(\frac{\alpha\beta\gamma + \alpha\gamma\delta + \alpha\beta\delta + \beta\gamma\delta}{\alpha\beta\gamma\delta}) = -(\frac{\Sigma \alpha\beta\gamma}{\alpha\beta\gamma\delta})$. Substituting the values: $-(\frac{-6}{-12}) = -(\frac{1}{2}) = -\frac{1}{2}$.

If $\alpha, \beta, \gamma$ and $\delta$ are zeroes of the polynomial equation $x^4-3x^2+6x-12=0$,then the value of $\frac{\alpha+\beta+\gamma}{\delta^2}+\frac{\alpha+\delta+\gamma}{\beta^2}+\frac{\alpha+\beta+\delta}{\gamma^2}+\frac{\delta+\beta+\gamma}{\alpha^2}$ is equal to

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