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

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(A) Given the cubic equation: ${x^3} + px + q = 0$ .....$(i)$From the properties of roots of a cubic equation $ax^3 + bx^2 + cx + d = 0$:Sum of roots:

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$-3q$

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(A) Given the cubic equation: ${x^3} + px + q = 0$ .....$(i)$From the properties of roots of a cubic equation $ax^3 + bx^2 + cx + d = 0$:Sum of roots: $\alpha + \beta + \gamma = -\frac{b}{a} = -\frac{0}{1} = 0$Sum of roots taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = p$Product of roots: $\alpha\beta\gamma = -\frac{d}{a} = -q$We use the algebraic identity: ${\alpha^3} + {\beta^3} + {\gamma^3} - 3\alpha\beta\gamma = (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha)$Since $\alpha + \beta + \gamma = 0$,the right side of the identity becomes $0$.Therefore,${\alpha^3} + {\beta^3} + {\gamma^3} - 3\alpha\beta\gamma = 0$${\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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