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

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(A) Given the cubic equation $x^3-x+1=0$. Let $f(x) = x^3-x+1$. By Vieta's formulas,for roots $\alpha, \beta, \gamma$: $\alpha+\beta+\gamma = 0$ $\alp

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(A) Given the cubic equation $x^3-x+1=0$. Let $f(x) = x^3-x+1$. By Vieta's formulas,for roots $\alpha, \beta, \gamma$: $\alpha+\beta+\gamma = 0$ $\alpha\beta+\beta\gamma+\gamma\alpha = -1$ $\alpha\beta\gamma = -1$ Let $y = \frac{1+x}{1-x}$. Then $y(1-x) = 1+x \implies y-yx = 1+x \implies y-1 = x(1+y) \implies x = \frac{y-1}{y+1}$. Since $x$ is a root of $x^3-x+1=0$,we substitute $x = \frac{y-1}{y+1}$: $(\frac{y-1}{y+1})^3 - (\frac{y-1}{y+1}) + 1 = 0$ $(y-1)^3 - (y-1)(y+1)^2 + (y+1)^3 = 0$ $(y^3-3y^2+3y-1) - (y-1)(y^2+2y+1) + (y^3+3y^2+3y+1) = 0$ $(y^3-3y^2+3y-1) - (y^3+2y^2+y-y^2-2y-1) + (y^3+3y^2+3y+1) = 0$ $y^3-3y^2+3y-1 - y^3-y^2+y+1 + y^3+3y^2+3y+1 = 0$ $y^3-y^2+7y+1 = 0$ The roots of this equation are $y_1 = \frac{1+\alpha}{1-\alpha}, y_2 = \frac{1+\beta}{1-\beta}, y_3 = \frac{1+\gamma}{1-\gamma}$. The sum of the roots is $y_1+y_2+y_3 = -(\frac{-1}{1}) = 1$.

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

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