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If $α, β$ are the roots of the equations $x^2 +px+1=0$, and $γ, δ$ are the roots of the equations $x^2+qx+1=0$, then $( α-γ)(β+ δ)( δ+ α)( β- γ) = $ ?

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    Compute away and use Viete. – 2017-01-08
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    @franzlemmermeyer viete? – 2017-01-08
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    Viete in French, Vieta in Latin. Or ask yourself what $\alpha + \beta$ and $\alpha \cdot \beta$ have to do with your first equation. – 2017-01-08

2 Answers 2

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We have $$P= (α - γ)(β + δ) (β - γ)(α + δ) = (αβ + αδ - βγ - γδ) (αβ + βδ - αγ - γδ)$$

We know that $αβ = γδ = 1$. Using these, we'll have

$$P = (1 + αδ - βγ - 1) (1 + βδ - αγ - 1)$$ $$ = (αδ - βγ) (βδ - αγ)$$ $$= αβδ^2 - γδα^2 - γδβ^2 + αβγ^2$$

Again, using $αβ = γδ = 1$, we get,

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

Now, 'completing the square', we have,

$$P = [(γ + δ)^2 - 2γδ] - [(α + β)^2 - 2αβ]$$

And using $α + β = -p , γ + δ = -q$ and $αβ = γδ = 1$, we get,

$$P = (q^2 - 2) - (p^2 - 2)= q^2 - p^2$$ Hope it helps.

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    Thank you @SteveAlsien. Happy it helped. – 2017-01-08
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    A small error: $a^2+b^2=(a+b)^2-2ab$ – 2017-01-08
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    @N74 Thank you for pointing out the typo. – 2017-01-08
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Write the product as $(\gamma - \alpha)(\gamma - \beta)(-\delta - \alpha)(-\delta - \beta) = P(\gamma) P(-\delta)$

$P(\gamma) = x^2+p \gamma+1 = p\gamma - q\gamma = \gamma (p-q)$ since $x^2+1 + q \gamma = 0$

Similarly $P(-\delta) = (q+p) \delta$

Hence $P(\gamma) P(\delta) = (q^2-p^2)\gamma \delta = q^2-p^2 $