If alpha, beta, gamma are the roots of x^3-2x | vedclass

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Question

(C) Given the equation $x^3-2x^2+4x-1=0$,the roots are $\alpha, \beta, \gamma$. From Vieta's formulas,we have $\alpha+\beta+\gamma=2$,$\alpha\beta+\be

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$x^3-8x^2+8x-8=0$

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(C) Given the equation $x^3-2x^2+4x-1=0$,the roots are $\alpha, \beta, \gamma$. From Vieta's formulas,we have $\alpha+\beta+\gamma=2$,$\alpha\beta+\beta\gamma+\gamma\alpha=4$,and $\alpha\beta\gamma=1$. Since $\alpha\beta\gamma=1$,we have $\beta\gamma=\frac{1}{\alpha}$,$\alpha\gamma=\frac{1}{\beta}$,and $\alpha\beta=\frac{1}{\gamma}$. The roots of the new equation are $\beta\gamma+\frac{1}{\alpha} = \frac{1}{\alpha}+\frac{1}{\alpha} = \frac{2}{\alpha}$,$\alpha\gamma+\frac{1}{\beta} = \frac{2}{\beta}$,and $\alpha\beta+\frac{1}{\gamma} = \frac{2}{\gamma}$. Let $y = \frac{2}{x}$,then $x = \frac{2}{y}$. Substituting this into the original equation: $(\frac{2}{y})^3 - 2(\frac{2}{y})^2 + 4(\frac{2}{y}) - 1 = 0$. $\frac{8}{y^3} - \frac{8}{y^2} + \frac{8}{y} - 1 = 0$. Multiplying by $-y^3$,we get $y^3 - 8y^2 + 8y - 8 = 0$. Thus,the required equation is $x^3-8x^2+8x-8=0$.

If $\alpha, \beta, \gamma$ are the roots of $x^3-2x^2+4x-1=0$,then the equation having the roots $\beta\gamma+\frac{1}{\alpha}, \alpha\beta+\frac{1}{\gamma}, \gamma\alpha+\frac{1}{\beta}$ is

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