If alpha, beta, gamma are the roots of the eq | vedclass

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(D) Let $f(x) = x^3 - x - 1$. Given that $\alpha, \beta, \gamma$ are the roots of $f(x) = 0$.Let $y = \frac{1+x}{1-x}$.Then $y(1-x) = 1+x \implies y -

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$-1$

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(D) Let $f(x) = x^3 - x - 1$. Given that $\alpha, \beta, \gamma$ are the roots of $f(x) = 0$.Let $y = \frac{1+x}{1-x}$.Then $y(1-x) = 1+x \implies y - yx = 1+x \implies y-1 = x(y+1) \implies x = \frac{y-1}{y+1}$.Substituting this value of $x$ into the equation $x^3 - x - 1 = 0$:$\left( \frac{y-1}{y+1} ight)^3 - \left( \frac{y-1}{y+1} ight) - 1 = 0$.Multiplying both sides by $(y+1)^3$:$(y-1)^3 - (y-1)(y+1)^2 - (y+1)^3 = 0$.Expanding the terms:$(y^3 - 3y^2 + 3y - 1) - (y-1)(y^2 + 2y + 1) - (y^3 + 3y^2 + 3y + 1) = 0$.$(y^3 - 3y^2 + 3y - 1) - (y^3 + y^2 - y - 1) - (y^3 + 3y^2 + 3y + 1) = 0$.$-y^3 - 7y^2 + y - 1 = 0 \implies y^3 + 7y^2 - y + 1 = 0$.The roots of this equation are $\frac{1+\alpha}{1-\alpha}, \frac{1+\beta}{1-\beta}, \frac{1+\gamma}{1-\gamma}$.The product of the roots is given by $-(	ext{constant term}) / (	ext{leading coefficient}) = -1/1 = -1$.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3 - x - 1 = 0$,then what is the value of $\left( \frac{1 + \alpha}{1 - \alpha} \right) \left( \frac{1 + \beta}{1 - \beta} \right) \left( \frac{1 + \gamma}{1 - \gamma} \right)$?

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