p is a non-zero real number. If the equation | vedclass

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(C) Let $\alpha, \beta, \gamma$ be the roots of the equation $x^3 - px^2 + px - 1 = 0$. From Vieta's formulas: $\alpha + \beta + \gamma = p$ $\alpha\b

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$3$

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(C) Let $\alpha, \beta, \gamma$ be the roots of the equation $x^3 - px^2 + px - 1 = 0$. From Vieta's formulas: $\alpha + \beta + \gamma = p$ $\alpha\beta + \beta\gamma + \gamma\alpha = p$ $\alpha\beta\gamma = 1$ Given that the equation with roots $\alpha^2, \beta^2, \gamma^2$ is identical to the original equation,the sum of these roots must also be $p$: $\alpha^2 + \beta^2 + \gamma^2 = p$ We know that $(\alpha + \beta + \gamma)^2 = \alpha^2 + \beta^2 + \gamma^2 + 2(\alpha\beta + \beta\gamma + \gamma\alpha)$. Substituting the known values: $p^2 = p + 2(p)$ $p^2 = 3p$ Since $p$ is non-zero,we divide by $p$: $p = 3$.

$p$ is a non-zero real number. If the equation whose roots are the squares of the roots of the equation $x^3 - px^2 + px - 1 = 0$ is identical to the given equation,then $p =$

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