If alpha, beta and gamma are roots of the equ | vedclass

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(B) Given that $\alpha, \beta, \gamma$ are roots of the cubic equation $x^3+4x-19=0$. Since $\alpha$ is a root,it must satisfy the equation: $\alpha^3

Vedclass · Accepted Answer

$3$

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(B) Given that $\alpha, \beta, \gamma$ are roots of the cubic equation $x^3+4x-19=0$. Since $\alpha$ is a root,it must satisfy the equation: $\alpha^3 + 4\alpha - 19 = 0$ $\Rightarrow \alpha^3 = 19 - 4\alpha$ Dividing both sides by $(19 - 4\alpha)$,we get: $\frac{\alpha^3}{19 - 4\alpha} = 1$ Similarly,since $\beta$ and $\gamma$ are also roots of the same equation: $\frac{\beta^3}{19 - 4\beta} = 1$ $\frac{\gamma^3}{19 - 4\gamma} = 1$ Adding these three expressions: $\frac{\alpha^3}{19-4\alpha} + \frac{\beta^3}{19-4\beta} + \frac{\gamma^3}{19-4\gamma} = 1 + 1 + 1 = 3$

If $\alpha, \beta$ and $\gamma$ are roots of the equation $x^3+4x-19=0$,then the value of $\frac{\alpha^3}{19-4\alpha}+\frac{\beta^3}{19-4\beta}+\frac{\gamma^3}{19-4\gamma}$ is equal to

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