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(B) Given the equation $x^3 + qx - r = 0$,the roots are $\alpha, \beta, \gamma$. From Vieta's formulas: $\alpha + \beta + \gamma = 0$ $\alpha \beta +

Vedclass · Accepted Answer

$rx^3 - q(r + 1)x^2 - (r + 1)^3 = 0$

Vedclass · Answer

(B) Given the equation $x^3 + qx - r = 0$,the roots are $\alpha, \beta, \gamma$. From Vieta's formulas: $\alpha + \beta + \gamma = 0$ $\alpha \beta + \beta \gamma + \gamma \alpha = q$ $\alpha \beta \gamma = r$ We need to find the equation with roots $y_1 = \beta \gamma + \frac{1}{\alpha}$,$y_2 = \gamma \alpha + \frac{1}{\beta}$,and $y_3 = \alpha \beta + \frac{1}{\gamma}$. Since $\alpha \beta \gamma = r$,we have $\beta \gamma = \frac{r}{\alpha}$,$\gamma \alpha = \frac{r}{\beta}$,and $\alpha \beta = \frac{r}{\gamma}$. Thus,the roots are $y_1 = \frac{r}{\alpha} + \frac{1}{\alpha} = \frac{r+1}{\alpha}$,$y_2 = \frac{r+1}{\beta}$,and $y_3 = \frac{r+1}{\gamma}$. Let $y = \frac{r+1}{x}$,then $x = \frac{r+1}{y}$. Substitute $x$ into the original equation: $(\frac{r+1}{y})^3 + q(\frac{r+1}{y}) - r = 0$ $\frac{(r+1)^3}{y^3} + \frac{q(r+1)}{y} - r = 0$ Multiply by $y^3$: $(r+1)^3 + q(r+1)y^2 - ry^3 = 0$ $ry^3 - q(r+1)y^2 - (r+1)^3 = 0$. Replacing $y$ with $x$,we get $rx^3 - q(r+1)x^2 - (r+1)^3 = 0$.

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

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