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(B) Given that $\alpha, \beta, \gamma$ are roots of $x^3+p x^2+q x+r=0$. From Vieta's formulas: $\alpha+\beta+\gamma = -p$,$\alpha\beta+\beta\gamma+\g

Vedclass · Accepted Answer

$q^2+p r$

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(B) Given that $\alpha, \beta, \gamma$ are roots of $x^3+p x^2+q x+r=0$. From Vieta's formulas: $\alpha+\beta+\gamma = -p$,$\alpha\beta+\beta\gamma+\gamma\alpha = q$,and $\alpha\beta\gamma = -r$. Let the new roots be $y_1 = \alpha(\beta+\gamma)$,$y_2 = \beta(\gamma+\alpha)$,and $y_3 = \gamma(\alpha+\beta)$. Note that $\alpha(\beta+\gamma) = \alpha(\beta+\gamma+\alpha) - \alpha^2 = -p\alpha - \alpha^2$. Since $\alpha^3+p\alpha^2+q\alpha+r=0$,we have $\alpha^3+p\alpha^2 = -q\alpha-r$. Dividing by $\alpha$,we get $\alpha^2+p\alpha = -q - \frac{r}{\alpha}$. Thus,$y_1 = -(-q - \frac{r}{\alpha}) = q + \frac{r}{\alpha}$. Similarly,$y_2 = q + \frac{r}{\beta}$ and $y_3 = q + \frac{r}{\gamma}$. Let $y = q + \frac{r}{x}$,then $x = \frac{r}{y-q}$. Substituting this into the original equation: $(\frac{r}{y-q})^3 + p(\frac{r}{y-q})^2 + q(\frac{r}{y-q}) + r = 0$. Dividing by $r$ (assuming $r 
eq 0$): $\frac{r^2}{(y-q)^3} + \frac{pr}{(y-q)^2} + \frac{q}{y-q} + 1 = 0$. $r^2 + pr(y-q) + q(y-q)^2 + (y-q)^3 = 0$. Expanding: $r^2 + pry - prq + q(y^2-2qy+q^2) + (y^3-3y^2q+3yq^2-q^3) = 0$. $y^3 + (q-3q)y^2 + (3q^2-2q^2+pr)y + (r^2-prq+q^3-q^3) = 0$. $y^3 - 2qy^2 + (q^2+pr)y + (r^2-prq) = 0$. The coefficient of $y$ is $q^2+pr$.

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3+p x^2+q x+r=0$,then the coefficient of $x$ in the cubic equation whose roots are $\alpha(\beta+\gamma), \beta(\gamma+\alpha)$ and $\gamma(\alpha+\beta)$ is

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