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(C) Given the cubic equation $5x^3-4x^2+3x-2=0$. Let the roots be $\alpha, \beta, \gamma$. From Vieta's formulas: $e_1 = \alpha+\beta+\gamma = -(\frac

Vedclass · Accepted Answer

$\frac{34}{125}$

Vedclass · Answer

(C) Given the cubic equation $5x^3-4x^2+3x-2=0$. Let the roots be $\alpha, \beta, \gamma$. From Vieta's formulas: $e_1 = \alpha+\beta+\gamma = -(\frac{-4}{5}) = \frac{4}{5}$ $e_2 = \alpha\beta+\beta\gamma+\gamma\alpha = \frac{3}{5}$ $e_3 = \alpha\beta\gamma = -(\frac{-2}{5}) = \frac{2}{5}$ Since $\alpha, \beta, \gamma$ are roots,they satisfy $5x^3 = 4x^2-3x+2$,or $x^3 = \frac{4}{5}x^2-\frac{3}{5}x+\frac{2}{5}$. Summing for all roots: $\sum \alpha^3 = \frac{4}{5}(\sum \alpha^2) - \frac{3}{5}(\sum \alpha) + 3(\frac{2}{5})$ We know $\sum \alpha^2 = (\sum \alpha)^2 - 2(\sum \alpha\beta) = (\frac{4}{5})^2 - 2(\frac{3}{5}) = \frac{16}{25} - \frac{6}{5} = \frac{16-30}{25} = -\frac{14}{25}$. Substituting back: $\sum \alpha^3 = \frac{4}{5}(-\frac{14}{25}) - \frac{3}{5}(\frac{4}{5}) + \frac{6}{5} = -\frac{56}{125} - \frac{12}{25} + \frac{6}{5} = \frac{-56-60+150}{125} = \frac{34}{125}$.

List-$I$	List-$II$
$(i) \alpha = \beta$	$(A) (ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$
$(ii) \alpha = 2\beta$	$(B) 2b^2 = 9ac$
$(iii) \alpha = 3\beta$	$(C) b^2 = 6ac$
$(iv) \alpha = \beta^2$	$(D) 3b^2 = 16ac$
	$(E) b^2 = 4ac$
	$(F) (ac^2)^{1/3} + (a^2c)^{1/3} = b$

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $5x^3-4x^2+3x-2=0$,then find the value of $\alpha^3+\beta^3+\gamma^3$.

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