Explanation

(a)
Here, $\alpha + \beta = - \frac{m}{l},\ \alpha\beta = \frac{n}{l}$
Now, $\alpha^{3}\beta + \alpha\beta^{3} = \alpha\beta\left( \alpha^{2} + \beta^{2} \right)$
$= \alpha\beta\left\lbrack (\alpha + \beta)^{2} - 2\alpha\beta \right\rbrack$
$= \frac{n}{l}\left\lbrack \left( \frac{- m}{l} \right)^{2} - \frac{2n}{l} \right\rbrack$
$= \frac{n}{l}\left( \frac{m^{2}}{l^{2}\ } - \frac{2n}{l} \right)$
And $\alpha^{3}\beta.\alpha\beta^{3} = (\alpha\beta)^{4} = \frac{n^{4}}{l^{4}}$
$\therefore$Required quadratic equation is
$x^{2} - \frac{n}{l}\left( \frac{m^{2}}{l^{2}} - \frac{2n}{l} \right)x + \frac{n^{4}}{l^{4}} = 0$
$\Rightarrow l^{4}x^{2} - nl\left( m^{2} - 2nl \right)x + n^{4} = 0$