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(B) Given,$\alpha, \beta$ and $\gamma$ are roots of the cubic equation $x^3-3x^2+x+5=0$. From the relation between roots and coefficients: $\alpha+\be

Vedclass · Accepted Answer

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

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(B) Given,$\alpha, \beta$ and $\gamma$ are roots of the cubic equation $x^3-3x^2+x+5=0$. From the relation between roots and coefficients: $\alpha+\beta+\gamma = -(-3)/1 = 3$ $\alpha \beta+\beta \gamma+\gamma \alpha = 1/1 = 1$ $\alpha \beta \gamma = -5/1 = -5$ Now,$y = \Sigma \alpha^2 + \alpha \beta \gamma = (\alpha^2+\beta^2+\gamma^2) + \alpha \beta \gamma$. Using the identity $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha \beta+\beta \gamma+\gamma \alpha)$: $y = (3)^2 - 2(1) + (-5) = 9 - 2 - 5 = 2$. 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 $y=\Sigma \alpha^2+\alpha \beta \gamma$ satisfies the equation

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