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(A) Given the cubic equation $x^3-ax^2+bx-c=0$ with roots $\alpha, \beta, \gamma$. By Vieta's formulas: $\alpha+\beta+\gamma = a$ $\alpha\beta+\beta\g

Vedclass · Accepted Answer

$ab-3c$

Vedclass · Answer

(A) Given the cubic equation $x^3-ax^2+bx-c=0$ with roots $\alpha, \beta, \gamma$. By Vieta's formulas: $\alpha+\beta+\gamma = a$ $\alpha\beta+\beta\gamma+\gamma\alpha = b$ $\alpha\beta\gamma = c$ We need to evaluate $\Sigma \alpha^2(\beta+\gamma)$. Since $\alpha+\beta+\gamma = a$,we have $\beta+\gamma = a-\alpha$. Substituting this into the expression: $\Sigma \alpha^2(a-\alpha) = a\Sigma \alpha^2 - \Sigma \alpha^3$ We know that $\Sigma \alpha^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = a^2-2b$. Also,for a cubic equation $x^3-ax^2+bx-c=0$,the sum of cubes $\Sigma \alpha^3$ is given by the identity: $\Sigma \alpha^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\Sigma \alpha^2 - \Sigma \alpha\beta)$ $\Sigma \alpha^3 = a(a^2-2b-b) + 3c = a(a^2-3b) + 3c = a^3-3ab+3c$. Now,substituting these into the expression: $a(a^2-2b) - (a^3-3ab+3c) = a^3-2ab-a^3+3ab-3c = ab-3c$. Thus,the correct option is $A$.

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

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