Question

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3-3 x^2-x-1=0$ and $f(x)=x-\frac{1}{x^2}-2$ then find the value of $f(\alpha) \cdot f(\beta) \cdot f(\gamma)$.

Answer 1

$\Rightarrow \alpha^3-3 \alpha^2-\alpha-1=0 $
$\Rightarrow \alpha^3-2 \alpha^2-1=\alpha^2+\alpha$
$\Rightarrow \alpha-2-\frac{1}{\alpha^2}=1+\frac{1}{\alpha} $
$\therefore f (\alpha) \cdot f (\beta) \cdot f (\gamma)=\left(1+\frac{1}{\alpha}\right)\left(1+\frac{1}{\beta}\right)\left(1+\frac{1}{\gamma}\right)$
$=1+\left(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\right)+\left(\frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\gamma \alpha}\right)+\frac{1}{\alpha \beta \gamma} $
$=1+\frac{\Sigma \alpha \beta}{\alpha \beta \gamma}+\frac{\Sigma \alpha}{\alpha \beta \gamma}+\frac{1}{\alpha \beta \gamma} $
$=1+\left(\frac{-1}{1}\right)+\frac{3}{1}+\frac{1}{1}=4$