If $\alpha$ and $\beta$ are the roots of the equation
x² + px + 2 = 0 and ${1 \over \alpha }$ and ${1 \over \beta }$ are the
roots of the equation 2x² + 2qx + 1 = 0, then
$$\left( {\alpha - {1 \over \alpha }} \right)\left( {\beta - {1 \over \beta }} \right)\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right)$$ is equal to :

$\alpha$ and $\beta$ are the roots of the <br><br>equation x² + px + 2 = 0 <br><br>$\therefore$ $\alpha + \beta = - p,\,\alpha \beta = 2$ <br><br>${1 \over \alpha }$ and ${1 \over \beta }$ are the roots of the <br><br>equation 2x² + 2qx + 1 = 0 <br><br>$\therefore$ $${1 \over \alpha } + {1 \over \beta } = - q,\,{1 \over {\alpha \beta }} = {1 \over 2}$$ <br><br>$\Rightarrow$ $${{\alpha + \beta } \over {\alpha \beta }} = - q \Rightarrow {{ - p} \over 2} = - q$$<br><br>$\Rightarrow p = 2q$<br><br>$$\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right) = \alpha \beta + {1 \over {\alpha \beta }} + 2$$ $= 2 + {1 \over 2} + 2 = {9 \over 2}$<br><br>$$\left( {\alpha - {1 \over \alpha }} \right)\left( {\beta - {1 \over \beta }} \right) = \alpha \beta + {1 \over {\alpha \beta }} - {\alpha \over \beta } - {\beta \over \alpha }$$<br><br>$$ = 2 + {1 \over 2} - \left[ {{{{\alpha ^2} + {\beta ^2}} \over {\alpha \beta }}} \right]$$$$<br><br>= {5 \over 2} - \left[ {{{{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta } \over {\alpha \beta }}} \right]$$<br><br>$= {5 \over 2} - \left[ {{{{p^2} - 4} \over 2}} \right]$<br><br>$= {{9 - {p^2}} \over 2}$<br><br>$$\left( {\alpha - {1 \over \alpha }} \right)\left( {\beta - {1 \over \beta }} \right)\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right) = \left( {{{9 - {p^2}} \over 2}} \right)\left( {{9 \over 2}} \right)$$<br><br>$= {9 \over 4}(9 - {p^2})$