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

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Question

(A) Given that $\alpha, \beta, \gamma$ are the roots of $x^3 - x - 1 = 0$.From the properties of roots,$\alpha + \beta + \gamma = 0$.Therefore,$\beta

Vedclass · Accepted Answer

$x^3 - x^2 + 1 = 0$

Vedclass · Answer

(A) Given that $\alpha, \beta, \gamma$ are the roots of $x^3 - x - 1 = 0$.From the properties of roots,$\alpha + \beta + \gamma = 0$.Therefore,$\beta + \gamma = -\alpha$,$\gamma + \alpha = -\beta$,and $\alpha + \beta = -\gamma$.The roots of the required equation are $\frac{1}{-\alpha}, \frac{1}{-\beta}, \frac{1}{-\gamma}$,which simplifies to $-\frac{1}{\alpha}, -\frac{1}{\beta}, -\frac{1}{\gamma}$.Let $y = -\frac{1}{x}$,so $x = -\frac{1}{y}$.Substitute $x = -\frac{1}{y}$ into the original equation $x^3 - x - 1 = 0$:$(-\frac{1}{y})^3 - (-\frac{1}{y}) - 1 = 0$$-\frac{1}{y^3} + \frac{1}{y} - 1 = 0$Multiply by $-y^3$:$1 - y^2 + y^3 = 0$$y^3 - y^2 + 1 = 0$.Thus,the required equation is $x^3 - x^2 + 1 = 0$.

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

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