If the cubic equation x^3-ax^2+ax-1=0 is iden | vedclass

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

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$3$

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(C) Let $\alpha, \beta, \gamma$ be the roots of the equation $x^3-ax^2+ax-1=0$. \ By Vieta's formulas: \ $\alpha+\beta+\gamma = a$ \ $\alpha\beta+\beta\gamma+\gamma\alpha = a$ \ $\alpha\beta\gamma = 1$ (Note: The constant term is $-1$,so product is $-(-1)/1 = 1$). \ The cubic equation whose roots are $\alpha^2, \beta^2, \gamma^2$ is $x^3 - (\alpha^2+\beta^2+\gamma^2)x^2 + (\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2)x - (\alpha\beta\gamma)^2 = 0$. \ Comparing this with $x^3-ax^2+ax-1=0$,we get: \ $a = \alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = a^2 - 2a$. \ Thus,$a^2 - 3a = 0$,which implies $a(a-3) = 0$. \ Since $a$ is non-zero,$a = 3$.

If the cubic equation $x^3-ax^2+ax-1=0$ is identical with the cubic equation whose roots are the squares of the roots of the given cubic equation,then the non-zero real value of $a$ is

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