If alpha, beta, gamma are the roots of the eq | vedclass

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(C) Given that $\alpha, \beta, \gamma$ are the roots of the equation $4x^3-3x^2+2x-1=0$. Since they are roots,they satisfy the equation: $4\alpha^3-3\

Vedclass · Accepted Answer

$\frac{3}{64}$

Vedclass · Answer

(C) Given that $\alpha, \beta, \gamma$ are the roots of the equation $4x^3-3x^2+2x-1=0$. Since they are roots,they satisfy the equation: $4\alpha^3-3\alpha^2+2\alpha-1=0 \Rightarrow 4\alpha^3=3\alpha^2-2\alpha+1$ $4\beta^3-3\beta^2+2\beta-1=0 \Rightarrow 4\beta^3=3\beta^2-2\beta+1$ $4\gamma^3-3\gamma^2+2\gamma-1=0 \Rightarrow 4\gamma^3=3\gamma^2-2\gamma+1$ Summing these equations: $4(\alpha^3+\beta^3+\gamma^3) = 3(\alpha^2+\beta^2+\gamma^2) - 2(\alpha+\beta+\gamma) + 3$ From Vieta's formulas: $\alpha+\beta+\gamma = \frac{3}{4}$,$\alpha\beta+\beta\gamma+\gamma\alpha = \frac{2}{4} = \frac{1}{2}$,$\alpha\beta\gamma = \frac{1}{4}$ We know $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = (\frac{3}{4})^2 - 2(\frac{1}{2}) = \frac{9}{16} - 1 = -\frac{7}{16}$ Substituting these values: $4(\alpha^3+\beta^3+\gamma^3) = 3(-\frac{7}{16}) - 2(\frac{3}{4}) + 3 = -\frac{21}{16} - \frac{3}{2} + 3 = -\frac{21}{16} + \frac{3}{2} = \frac{-21+24}{16} = \frac{3}{16}$ Therefore,$\alpha^3+\beta^3+\gamma^3 = \frac{3}{64}$.

If $\alpha, \beta, \gamma$ are the roots of the equation $4x^3-3x^2+2x-1=0$,then $\alpha^3+\beta^3+\gamma^3=$

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