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(D) Since $\alpha, \beta, \gamma$ are roots of $x^3 - x - 2 = 0$,we have $\alpha^3 = \alpha + 2$,$\beta^3 = \beta + 2$,and $\gamma^3 = \gamma + 2$.Mul

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(D) Since $\alpha, \beta, \gamma$ are roots of $x^3 - x - 2 = 0$,we have $\alpha^3 = \alpha + 2$,$\beta^3 = \beta + 2$,and $\gamma^3 = \gamma + 2$.Multiplying by $\alpha^2$,we get $\alpha^5 = \alpha^3 + 2\alpha^2 = (\alpha + 2) + 2\alpha^2 = 2\alpha^2 + \alpha + 2$.Similarly,$\beta^5 = 2\beta^2 + \beta + 2$ and $\gamma^5 = 2\gamma^2 + \gamma + 2$.Summing these,$\alpha^5 + \beta^5 + \gamma^5 = 2(\alpha^2 + \beta^2 + \gamma^2) + (\alpha + \beta + \gamma) + 6$.From the equation $x^3 + 0x^2 - x - 2 = 0$,we have $\sum \alpha = 0$ and $\sum \alpha\beta = -1$.Using the identity $\alpha^2 + \beta^2 + \gamma^2 = (\sum \alpha)^2 - 2(\sum \alpha\beta) = 0^2 - 2(-1) = 2$.Substituting these values: $\alpha^5 + \beta^5 + \gamma^5 = 2(2) + 0 + 6 = 10$.

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

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