If α, β are the roots of the equation x

Solution:

(a) $q^{2}-p^{2}$

Given: $\alpha$ and $\beta$ are the roots of the equation $x^{2}+p x+1=0$.

Also, $\gamma$ and $\delta$ are the roots of the equation $x^{2}+q x+1=0$.

Then, the sum and the product of the roots of the given equation are as follows:

$\alpha+\beta=-\frac{p}{1}=-p$

$\alpha \beta=\frac{1}{1}=1$

$\gamma+\delta=-\frac{q}{1}=-q$

$\gamma \delta=\frac{1}{1}=1$

Moreover, $(\gamma+\delta)^{2}=\gamma^{2}+\delta^{2}+2 \gamma \delta$

$\Rightarrow \gamma^{2}+\delta^{2}=q^{2}-2$

$\therefore(\alpha-\gamma)(\alpha+\delta)(\beta-\gamma)(\beta+\delta)=(\alpha-\gamma)(\beta-\gamma)(\alpha+\delta)(\beta+\delta)$

$=\left(\alpha \beta-\alpha \gamma-\beta \gamma+\gamma^{2}\right)\left(\alpha \beta+\alpha \delta+\beta \delta+\delta^{2}\right)$

$=\left[\alpha \beta-\gamma(\alpha+\beta)+\gamma^{2}\right]\left[\alpha \beta+\delta(\alpha+\beta)+\delta^{2}\right]$

$=\left(1-\gamma(-p)+\gamma^{2}\right)\left(1+\delta(-p)+\delta^{2}\right)$

$=\left(1+\gamma p+\gamma^{2}\right)\left(1-\delta p+\delta^{2}\right)$

$=1-p \delta+\delta^{2}+p \gamma-p^{2} \gamma \delta+p \gamma \delta^{2}+\gamma^{2}-p \delta \gamma^{2}+\gamma^{2} \delta^{2}$

$=1-p \delta+p \gamma+\delta^{2}-p^{2} \gamma \delta+p \gamma \delta^{2}+\gamma^{2}-p \delta \gamma^{2}+\gamma^{2} \delta^{2}$

$=1-p(\delta-\gamma)-p^{2} \gamma \delta+p \gamma \delta(\delta-\gamma)+\left(\gamma^{2}+\delta^{2}\right)+1$

$=1-p^{2} \gamma \delta+p \gamma \delta(\delta-\gamma)-p(\delta-\gamma)+\left(\gamma^{2}+\delta^{2}\right)+1$

$=1-p^{2}+(\delta-\gamma) p(\gamma \delta-1)+q^{2}-2+1$

$=-p^{2}+(\delta-\gamma) p(1-1)+q^{2}$

$=q^{2}-p^{2}$