If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-a x^2+b x-c=0$, then $\Sigma \alpha^2(\beta+\gamma)=$

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Aicharya · Accepted Answer

If $\alpha, \beta, \gamma$ are the roots of the equation. $ x^3-a x^2+b x-c=0 $ Then, $\alpha+\beta+\gamma=a, \alpha \beta+\beta \gamma+\gamma \alpha=b$ and $\alpha \beta \gamma=c$. Now, $\Sigma \alpha^2(\beta+\gamma)=\Sigma \alpha^2(a-\alpha)$ [as $\alpha+\beta+\gamma=a]$ $ \begin{aligned} & =a \sum \alpha^2-\sum \alpha^3 \ & =a\left[(\alpha+\beta+\gamma)^2-2(\alpha \beta+\beta \gamma+\gamma \alpha)ight]-[(\alpha+\beta+\gamma) \ & \left.\quad\left(\alpha^2+\beta^2+\gamma^2-\alpha \beta-\beta \gamma-\gamma \alphaight)+3 \alpha \beta \gammaight] \ & =a\left[(\alpha+\beta+\gamma)^2-2(\alpha \beta+\beta \gamma+\gamma \alpha)ight]-(\alpha+\beta+\gamma) \ & \quad\left\{(\alpha+\beta+\gamma)^2-3(\alpha \beta+\beta \gamma+\gamma \alpha)+3 \alpha \beta \gammaight] \ & =a\left[a^2-2 bight]-\left[a\left\{a^2-3 bight\}+3 cight] \ & =a^3-2 a b-a^3+3 a b-3 c=a b-3 c \end{aligned} $ Hence, option (a) is correct

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-a x^2+b x-c=0$, then $\Sigma…

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