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

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(A) Given the cubic equation $x^3 + px + q = 0$. Since $\alpha, \beta, \gamma$ are the roots,by Vieta's formulas: $\alpha + \beta + \gamma = 0$ (coeff

Vedclass · Accepted Answer

$-3q$

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(A) Given the cubic equation $x^3 + px + q = 0$. Since $\alpha, \beta, \gamma$ are the roots,by Vieta's formulas: $\alpha + \beta + \gamma = 0$ (coefficient of $x^2$ is $0$). $\alpha\beta + \beta\gamma + \gamma\alpha = p$. $\alpha\beta\gamma = -q$. We know the algebraic identity: $\alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma = (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha)$. Since $\alpha + \beta + \gamma = 0$,the right side of the identity becomes $0$. Therefore,$\alpha^3 + \beta^3 + \gamma^3 = 3\alpha\beta\gamma$. Substituting $\alpha\beta\gamma = -q$,we get $\alpha^3 + \beta^3 + \gamma^3 = 3(-q) = -3q$.

If $\alpha, \beta$ and $\gamma$ are the roots of $x^3 + px + q = 0$,then the value of $\alpha^3 + \beta^3 + \gamma^3$ is equal to

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