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$p$ is non-zero real number. If the equation whose roots are the squares of the roots of the equation $x^3-p x^2+p x-1=0$ is identical with the given equation, then $p=$
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The correct answer is:
3
Let $\alpha, \beta, \gamma$ are roots of equation
$x^3-p x^2+p x-1=0$
$\because \quad \alpha+\beta+\gamma=p$
$\alpha \beta+\beta \gamma+\alpha \gamma=p$
$\alpha \beta \gamma=1$
Also given, $\quad \alpha^2+\beta^2+\gamma^2=p$
$\alpha^2 \beta^2+\beta^2 \gamma^2+\alpha^2 \gamma^2=p$
$\alpha^2 \beta^2 \gamma^2=1$
$\therefore(\alpha+\beta+\gamma)^2=\alpha^2+\beta^2+\gamma^2+2(\alpha \beta+\beta \gamma+\gamma \alpha)$
$p^2=p+2 p$
$p^2=3 p$
$p=3$
$x^3-p x^2+p x-1=0$
$\because \quad \alpha+\beta+\gamma=p$
$\alpha \beta+\beta \gamma+\alpha \gamma=p$
$\alpha \beta \gamma=1$
Also given, $\quad \alpha^2+\beta^2+\gamma^2=p$
$\alpha^2 \beta^2+\beta^2 \gamma^2+\alpha^2 \gamma^2=p$
$\alpha^2 \beta^2 \gamma^2=1$
$\therefore(\alpha+\beta+\gamma)^2=\alpha^2+\beta^2+\gamma^2+2(\alpha \beta+\beta \gamma+\gamma \alpha)$
$p^2=p+2 p$
$p^2=3 p$
$p=3$
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