$\alpha, \beta, \gamma$ are the roots of the equation $x^3+3 x^2-10 x-24=$ 0 . If $\alpha(\beta+\gamma), \beta(\gamma+\alpha)$ and $\gamma(\alpha+\beta)$ are the roots of the equation $x^3+p x^2+q x+r=0$, then $q=$

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Aicharya · Accepted Answer

Since, $\alpha, \beta, \gamma$ be the roots of the equation $\begin{aligned} & x^3+3 x^2-10 x-24=0 \ & \because \alpha \beta+\beta \gamma+\alpha \gamma=-10, \alpha \beta \gamma=24, \alpha+\beta+\gamma=-3 \end{aligned}$ Now, $(\alpha \beta+\beta \gamma+\alpha \gamma)^2=\alpha^2 \beta^2+\beta^2 \gamma^2+\alpha^2 \gamma^2$ $\Rightarrow \alpha^2 \beta^2+\beta^2 \gamma^2+\alpha^2 \gamma^2=244 \quad+2 \alpha \beta \gamma(\alpha+\beta+\gamma)$ $\begin{aligned} & 	ext { Since, } q=\alpha \beta(\beta+\gamma)(\gamma+\alpha)+\beta \gamma(\gamma+\alpha)(\alpha+\beta) \ & +\alpha \gamma(\beta+\gamma)(\alpha+\beta) \ & =\alpha \beta \gamma(\alpha+\beta+\gamma)+\alpha^2 \beta^2+\alpha \beta \gamma(\alpha+\beta+\gamma) \ & +\beta^2 \gamma^2+\alpha \beta \gamma(\alpha+\beta+\gamma)+\alpha^2 \gamma^2 \ & =3 \cdot 24 \cdot(-3)+244=-216+244=28 . \end{aligned}$

$\alpha, \beta, \gamma$ are the roots of the equation $x^3+3 x^2-10 x-24=$ 0 . If $\alpha(\beta+\gamma),…

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