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(D) Given the equation: $x^3-ax^2+bx-c=0$ $(i)$.Let $\alpha, \beta, \gamma$ be the roots of equation $(i)$.From Vieta's formulas,we have:$\alpha+\beta

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$3$

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(D) Given the equation: $x^3-ax^2+bx-c=0$ $(i)$.Let $\alpha, \beta, \gamma$ be the roots of equation $(i)$.From Vieta's formulas,we have:$\alpha+\beta+\gamma=a$$\alpha\beta+\beta\gamma+\gamma\alpha=b$$\alpha\beta\gamma=c$Given that the sum of the cubes of the roots is zero: $\alpha^3+\beta^3+\gamma^3=0$.We use the identity: $\alpha^3+\beta^3+\gamma^3-3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha)$.Substituting the given sum: $0-3c = (\alpha+\beta+\gamma)((\alpha+\beta+\gamma)^2-3(\alpha\beta+\beta\gamma+\gamma\alpha))$.$-3c = a(a^2-3b)$.$-3c = a^3-3ab$.Rearranging the terms,we get: $a^3+3c = 3ab$.

If the sum of the cubes of the roots of the equation $x^3-ax^2+bx-c=0$ is zero,then $a^3+3c=$ (in $ab$)

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