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

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(A) Given the cubic equation $5x^3 - 2x - 4 = 0$. Since $\alpha, \beta, \gamma$ are the roots,they must satisfy the equation: $5\alpha^3 - 2\alpha - 4

Vedclass · Accepted Answer

$\frac{12}{5}$

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(A) Given the cubic equation $5x^3 - 2x - 4 = 0$. Since $\alpha, \beta, \gamma$ are the roots,they must satisfy the equation: $5\alpha^3 - 2\alpha - 4 = 0 \implies 5\alpha^3 = 2\alpha + 4$ $5\beta^3 - 2\beta - 4 = 0 \implies 5\beta^3 = 2\beta + 4$ $5\gamma^3 - 2\gamma - 4 = 0 \implies 5\gamma^3 = 2\gamma + 4$ Summing these equations: $5(\alpha^3 + \beta^3 + \gamma^3) = 2(\alpha + \beta + \gamma) + 12$ From the relation between roots and coefficients for $ax^3 + bx^2 + cx + d = 0$,the sum of roots $\alpha + \beta + \gamma = -\frac{b}{a}$. Here,$a = 5, b = 0$,so $\alpha + \beta + \gamma = 0$. Substituting this: $5(\alpha^3 + \beta^3 + \gamma^3) = 2(0) + 12 = 12$ $\alpha^3 + \beta^3 + \gamma^3 = \frac{12}{5}$

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

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