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

Vedclass · Accepted Answer

$q^2+p r$

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(B) Given $\alpha, \beta, \gamma$ are roots of $x^3+p x^2+q x+r=0$. By 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 $\alpha$ is a root,$\alpha^3+p\alpha^2+q\alpha+r=0$,so $\alpha^3+p\alpha^2 = -q\alpha-r$. Thus,$y_1 = -p\alpha - \alpha^2$. Alternatively,$y_1 = \alpha\beta+\alpha\gamma = q - \beta\gamma = q - (\frac{-r}{\alpha}) = q + \frac{r}{\alpha}$. So,$\alpha = \frac{r}{y-q}$. 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 $r 
eq 0$): $\frac{r^2}{(y-q)^3} + \frac{pr}{(y-q)^2} + \frac{q}{y-q} + 1 = 0$. Multiplying by $(y-q)^3$: $r^2 + pr(y-q) + q(y-q)^2 + (y-q)^3 = 0$. 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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