If the roots alpha, beta, gamma of the equati | vedclass

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(A) Given that $\alpha, \beta, \gamma$ are the roots of the equation $x^3-6x^2+px+10=0$ and are in arithmetic progression $(AP)$. Let the roots be $\b

Vedclass · Accepted Answer

$132$

Vedclass · Answer

(A) Given that $\alpha, \beta, \gamma$ are the roots of the equation $x^3-6x^2+px+10=0$ and are in arithmetic progression $(AP)$. Let the roots be $\beta-d, \beta, \beta+d$. From the sum of roots,$(\beta-d) + \beta + (\beta+d) = 6$,which gives $3\beta = 6$,so $\beta = 2$. Since $\beta = 2$ is a root,it must satisfy the equation: $2^3 - 6(2^2) + p(2) + 10 = 0$. $8 - 24 + 2p + 10 = 0$ $\Rightarrow 2p - 6 = 0$ $\Rightarrow p = 3$. The product of the roots is $\alpha\beta\gamma = -10$. Since $\beta = 2$,we have $\alpha\gamma = -5$. Also,$\alpha+\gamma = 6 - 2 = 4$. We use the identity $\alpha^3+\beta^3+\gamma^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2 - (\alpha\beta+\beta\gamma+\gamma\alpha))$. Alternatively,$\alpha^3+\beta^3+\gamma^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)((\alpha+\beta+\gamma)^2 - 3(\alpha\beta+\beta\gamma+\gamma\alpha))$. Here,$\alpha+\beta+\gamma = 6$,$\alpha\beta+\beta\gamma+\gamma\alpha = p = 3$,and $\alpha\beta\gamma = -10$. $\alpha^3+\beta^3+\gamma^3 - 3(-10) = 6(6^2 - 3(3))$. $\alpha^3+\beta^3+\gamma^3 + 30 = 6(36 - 9) = 6(27) = 162$. $\alpha^3+\beta^3+\gamma^3 = 162 - 30 = 132$.

If the roots $\alpha, \beta, \gamma$ of the equation $x^3-6x^2+px+10=0$ are in arithmetic progression,then $\alpha^3+\beta^3+\gamma^3=$

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