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

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Question

(A) Given the equation $x^3 - x - 1 = 0$ with roots $\alpha, \beta, \gamma$.From the properties of roots,the sum of roots is $\alpha + \beta + \gamma

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the equation $x^3 - x - 1 = 0$ with roots $\alpha, \beta, \gamma$.From the properties of roots,the sum of roots is $\alpha + \beta + \gamma = 0$.Therefore,we can write: $\beta + \gamma = -\alpha$,$\gamma + \alpha = -\beta$,and $\alpha + \beta = -\gamma$.The roots of the required equation are $\frac{1}{-\alpha}, \frac{1}{-\beta}, \frac{1}{-\gamma}$,which simplifies to $-\frac{1}{\alpha}, -\frac{1}{\beta}, -\frac{1}{\gamma}$.To find the equation with roots $-\frac{1}{x}$,we substitute $x$ with $-x$ in the original equation to get $(-x)^3 - (-x) - 1 = 0$,which is $-x^3 + x - 1 = 0$,or $x^3 - x + 1 = 0$. Wait,let's re-evaluate: If the roots are $y = -\frac{1}{x}$,then $x = -\frac{1}{y}$.Substituting $x = -\frac{1}{y}$ into $x^3 - x - 1 = 0$:$(-\frac{1}{y})^3 - (-\frac{1}{y}) - 1 = 0$$-\frac{1}{y^3} + \frac{1}{y} - 1 = 0$Multiplying by $-y^3$: $1 - y^2 + y^3 = 0$,which is $y^3 - y^2 + 1 = 0$.

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

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