If alpha, beta, gamma are the roots of the eq | vedclass

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(B) Given the cubic equation $x^3+px^2+qx+r=0$ with roots $\alpha, \beta, \gamma$. By Vieta's formulas: $\alpha+\beta+\gamma = -p$ $\alpha\beta+\beta\

Vedclass · Accepted Answer

$r-pq$

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(B) Given the cubic equation $x^3+px^2+qx+r=0$ with roots $\alpha, \beta, \gamma$. By Vieta's formulas: $\alpha+\beta+\gamma = -p$ $\alpha\beta+\beta\gamma+\gamma\alpha = q$ $\alpha\beta\gamma = -r$ We know that $(\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha) = (\alpha+\beta)(\beta+\gamma)(\gamma+\alpha) + \alpha\beta\gamma$. Substituting the values: $(-p)(q) = (\alpha+\beta)(\beta+\gamma)(\gamma+\alpha) + (-r)$ $-pq = (\alpha+\beta)(\beta+\gamma)(\gamma+\alpha) - r$ Therefore,$(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha) = r-pq$.

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

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