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(C) Given the cubic equation $x^3-9x^2+23x-15=0$. By Vieta's formulas,for roots $\alpha, \beta, \gamma$: $\alpha+\beta+\gamma = 9$ ... $(i)$ $\alpha\b

Vedclass · Accepted Answer

$153$

Vedclass · Answer

(C) Given the cubic equation $x^3-9x^2+23x-15=0$. By Vieta's formulas,for roots $\alpha, \beta, \gamma$: $\alpha+\beta+\gamma = 9$ ... $(i)$ $\alpha\beta+\beta\gamma+\gamma\alpha = 23$ ... (ii) $\alpha\beta\gamma = 15$ ... (iii) Since $\alpha, \beta, \gamma$ are roots,they satisfy the equation: $\alpha^3-9\alpha^2+23\alpha-15=0 \implies \alpha^3 = 9\alpha^2-23\alpha+15$ $\beta^3-9\beta^2+23\beta-15=0 \implies \beta^3 = 9\beta^2-23\beta+15$ $\gamma^3-9\gamma^2+23\gamma-15=0 \implies \gamma^3 = 9\gamma^2-23\gamma+15$ Adding these three equations: $\alpha^3+\beta^3+\gamma^3 = 9(\alpha^2+\beta^2+\gamma^2) - 23(\alpha+\beta+\gamma) + 45$ We know $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = 9^2 - 2(23) = 81 - 46 = 35$. Substituting the values: $\alpha^3+\beta^3+\gamma^3 = 9(35) - 23(9) + 45$ $= 315 - 207 + 45 = 153$. Thus,the correct option is $C$.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-9x^2+23x-15=0$,then $\alpha^3+\beta^3+\gamma^3=$

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