α, β are roots of x^2 + px + 1 = 0 and γ, δ are roots of equation x^2 + qx + 1 = 0, show that (α - γ)(β - γ)(α + δ)(β + δ) = q^2 - r^2

x^2 + px + 1 = 0 has roots α, β

x^2 + qx + 1 = 0 has roots γ, δ

Using the sum and product of roots formulae, we'll get

α + β = -p ; αβ = 1
γ + δ = -q ; γδ = 1

We have to prove that 

(α - γ)(β - γ)(α + δ)(β + δ) = q^2 - r^2

To begin with Left Hand Side,

LHS

= (α - γ)(β - γ)(α + δ)(β + δ)

This is a product of four expressions. Direct simplification can be a lot difficult, and can even be useless as it won't be easy to draw anything useful from it. We will begin with multiplying expressions in pairs of two.

So, think about simplifying (α - γ)(β - γ) and (α + δ)(β + δ). Anything wrong in doing so? Nope, not at all. But doing so will give us two Quadratic expressions. The first one will give us (γ^2 + something + something) and the second one will give (δ^2 + something + something), and again, it is not going to be simple to convert them into what we want to see. Then what else can be done?

Think of a possible pairing that won't lead to expression in 2 degree. Its very simple. Yes, it is (α - γ)(β + δ) and (β - γ)(α + δ). Check them out carefully, and you'll understand why they don't lead to (something)^2.

= (α - γ)(β + δ) (β - γ)(α + δ)
= (αβ + αδ - βγ - γδ) (αβ + βδ - αγ - γδ)

We found out that αβ = γδ = 1. Using these, we'll have

= (1 + αδ - βγ - 1) (1 + βδ - αγ - 1)
= (αδ - βγ) (βδ - αγ)
= αβ.δ^2 - γδ.α^2 - γδ.β^2 + αβγ^2

Again, using αβ = γδ = 1

= δ^2 - α^2 - β^2 + γ^2
= γ^2 + δ^2 - (α^2 + β^2)

Now, 'completing the square' will be used

= [(γ + δ)^2 - γδ] - [(α + β)^2 - αβ]

Using α + β = -p , γ + δ = -q , αβ = γδ = 1 

= (q^2 - 1) - (p^2 - 1)
= q^2 - p^2 = RHS

Hence Proved.