If alpha, beta, gamma are the roots of x^3+p | vedclass

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(A) Given the cubic equation $x^3+p x^2+q x+r=0$ with roots $\alpha, \beta, \gamma$.From Vieta's formulas:$\alpha+\beta+\gamma = -p$$\alpha \beta+\bet

Vedclass · Accepted Answer

$3 p q-3 r-p^3$

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(A) Given the cubic equation $x^3+p x^2+q x+r=0$ with roots $\alpha, \beta, \gamma$.From Vieta's formulas:$\alpha+\beta+\gamma = -p$$\alpha \beta+\beta \gamma+\gamma \alpha = q$$\alpha \beta \gamma = -r$We use the algebraic identity:$\alpha^3+\beta^3+\gamma^3-3 \alpha \beta \gamma = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2-\alpha \beta-\beta \gamma-\gamma \alpha)$Also,$\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha \beta+\beta \gamma+\gamma \alpha) = (-p)^2 - 2q = p^2-2q$.Substituting these into the identity:$\alpha^3+\beta^3+\gamma^3 - 3(-r) = (-p)(p^2-2q-q)$$\alpha^3+\beta^3+\gamma^3 + 3r = -p(p^2-3q)$$\alpha^3+\beta^3+\gamma^3 = -p^3+3pq-3r$Rearranging,we get $\alpha^3+\beta^3+\gamma^3 = 3pq-3r-p^3$.

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

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