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(B) Given the cubic equation $x^3 - 3x^2 + x + 5 = 0$.By Vieta's formulas,we have:$\alpha + \beta + \gamma = 3$$\alpha \beta + \beta \gamma + \gamma \

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$y^3 - y^2 - y - 2 = 0$

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(B) Given the cubic equation $x^3 - 3x^2 + x + 5 = 0$.By Vieta's formulas,we have:$\alpha + \beta + \gamma = 3$$\alpha \beta + \beta \gamma + \gamma \alpha = 1$$\alpha \beta \gamma = -5$We need to find the value of $y = \sum \alpha^2 + \alpha \beta \gamma$.Recall that $\sum \alpha^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha \beta + \beta \gamma + \gamma \alpha)$.Substituting the values:$\sum \alpha^2 = (3)^2 - 2(1) = 9 - 2 = 7$.Now,$y = 7 + (\alpha \beta \gamma) = 7 + (-5) = 2$.Since $y = 2$,we check which equation is satisfied by $y = 2$:For option $B$: $y^3 - y^2 - y - 2 = (2)^3 - (2)^2 - 2 - 2 = 8 - 4 - 2 - 2 = 0$.Thus,$y = 2$ satisfies the equation $y^3 - y^2 - y - 2 = 0$.

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3 - 3x^2 + x + 5 = 0$,then $y = \sum \alpha^2 + \alpha \beta \gamma$ satisfies which of the following equations?

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