Here is the question:
If the roots of the equation
$$
x^4 – px^3 + qx^2 – pqx + 1 = 0
$$
are $\alpha, \beta, \gamma,$ and $\delta$, show that
$$
(\alpha + \beta + \gamma)(\alpha + \beta + \delta)
(\alpha + \gamma + \delta)(\beta + \gamma + \delta)
= 1.
$$
Pretty much exhausted my resources.If there are more than one way of doing it please state and you can state some good books for this particular topic.
$$
1+x^4+\text{qx}^2-\text{px}^3-\text{pqx} \equiv x^4+x^3 (-\alpha -\beta -\gamma -\delta )+x^2
(\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta
+\gamma \delta )+x (-\alpha \beta \gamma -\alpha \beta \delta -\alpha
\gamma \delta -\beta \gamma \delta )+\alpha \beta \gamma \delta
$$
So,
$$
\text{p} = \alpha+\beta +\gamma +\delta
$$
$$
\text{q} = \alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta
+\gamma \delta
$$
$$
\text{p}\text{q}=\alpha \beta \gamma +\alpha \beta \delta +\alpha
\gamma \delta +\beta \gamma \delta
$$
$$
1= \alpha \beta \gamma \delta
$$
Best Answer
We know the factorisation:
$f(x) = x^4 - px^3 + qx^2 - pqx + 1 = (x - \alpha)(x - \beta)(x - \gamma)(x - \delta)$
Equating $x^3$ coefficients gives $\alpha + \beta + \gamma + \delta = p$.
Thus we can write the expression in question as:
$(p - \delta)(p - \gamma)(p - \beta)(p - \alpha) = f(p) = 1$.