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If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+3 x^2-x-3=0$, then $\left(1+\alpha^2\right)\left(1+\beta^2\right)\left(1+\gamma^2\right)=$
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40
Given equation, $x^3+3 x^2-x-3=0$
$\Rightarrow \quad x^2(x+3)-1(x+3)=0$
$\Rightarrow \quad(x-1)(x+1)(x+3)=0$
So, roots $\alpha=-3, \beta=-1$ and $\gamma=1$
Therefore,
$\left(1+\alpha^2\right)\left(1+\beta^2\right)\left(1+\gamma^2\right)$
$=(1+9)(1+1)(1+1)=40$
$\Rightarrow \quad x^2(x+3)-1(x+3)=0$
$\Rightarrow \quad(x-1)(x+1)(x+3)=0$
So, roots $\alpha=-3, \beta=-1$ and $\gamma=1$
Therefore,
$\left(1+\alpha^2\right)\left(1+\beta^2\right)\left(1+\gamma^2\right)$
$=(1+9)(1+1)(1+1)=40$
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