Question

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

Answer 1

Answer: 4

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