Explanation

(b)
Here, $\alpha + \beta = p\ and\ \alpha\beta = q$
Also, $\alpha_{1} + \beta_{1} = q\ and\ {\alpha_{1}\beta}_{1} = p$
$\therefore$ Sum of given roots
$= \left( \ \frac{1}{\alpha_{1}\beta} + \frac{1}{\alpha\beta_{1}} \right) + \left( \ \frac{1}{\alpha\alpha_{1}} + \frac{1}{\beta\beta_{1}} \right)$
$= \frac{\alpha\beta_{1} + \alpha_{1}\beta + \beta\beta_{1} + \alpha\alpha_{1}\ }{\alpha\beta\alpha_{1}\beta_{1}}$
$= \frac{(\alpha + \beta)(\alpha_{1} + \beta_{1})}{(\alpha\beta)(\alpha_{1}\beta_{1})} = \frac{pq}{qp} = 1$
and product of given roots
$= \left( \ \frac{1}{\alpha_{1}\beta} + \frac{1}{\alpha\beta_{1}} \right)\left( \ \frac{1}{\alpha\alpha_{1}} + \frac{1}{\beta\beta_{1}} \right)$
$= \frac{\left( \alpha\beta_{1} + \alpha_{1}\beta \right)(\alpha\alpha_{1} + \beta\beta_{1})}{\alpha^{2}\beta^{2}\alpha_{1}^{2}\beta_{1}^{2}}$
$= \frac{\alpha\beta\left( \alpha_{1 +}^{2}\beta_{1}^{2} \right) + {\alpha_{1}\beta}_{1}(\alpha^{2} + \beta^{2})}{\alpha^{2}\beta^{2}\alpha_{1}^{2}\beta_{1}^{2}}$
$= \frac{\begin{array}{r}
\alpha\beta\left\lbrack \left( \alpha_{1} + \beta_{1} \right)^{2} - 2\alpha_{1}\beta \right\rbrack + \\
\alpha_{1}\beta_{1}\lbrack(\alpha + \beta)^{2} - 2\alpha\beta\rbrack
\end{array}}{(\alpha\beta)^{2}\left( \alpha_{1}\beta_{1} \right)^{2}}$
$= \frac{q\left( q^{2} - 2p \right) + p(p^{2} - 2q)}{q^{2}p^{2}}$
$= \frac{p^{3} + q^{3} - 4qp}{p^{2}q^{2}}$
Hence, the required equation is given by
$x^{2} - (sum\ of\ roots)x + (product\ of\ roots) = 0$
$\Rightarrow \left( p^{2}q^{2} \right)x^{2} - \left( p^{2}q^{2} \right)x + p^{3} + q^{3} - 4qp = 0$