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Question

(B) Given the equation $x^3 + qx - r = 0$,we have $\alpha + \beta + \gamma = 0$,$\alpha \beta + \beta \gamma + \gamma \alpha = q$,and $\alpha \beta \g

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given the equation $x^3 + qx - r = 0$,we have $\alpha + \beta + \gamma = 0$,$\alpha \beta + \beta \gamma + \gamma \alpha = q$,and $\alpha \beta \gamma = r$.Since $\alpha \beta \gamma = r$,we have $\beta \gamma = \frac{r}{\alpha}$,$\gamma \alpha = \frac{r}{\beta}$,and $\alpha \beta = \frac{r}{\gamma}$.The roots of the new equation are $\frac{r}{\alpha} + \frac{1}{\alpha} = \frac{r+1}{\alpha}$,$\frac{r+1}{\beta}$,and $\frac{r+1}{\gamma}$.Let $y = \frac{r+1}{x}$,which implies $x = \frac{r+1}{y}$.Substituting this into the original equation: $\left( \frac{r+1}{y} \right)^3 + q \left( \frac{r+1}{y} \right) - r = 0$.Multiplying by $y^3$: $(r+1)^3 + q(r+1)y^2 - ry^3 = 0$.Rearranging gives $ry^3 - q(r+1)y^2 - (r+1)^3 = 0$.

If $\alpha, \beta, \gamma$ are 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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