Polynomials

$\large f(x)=x^2+px+1=(x-a)(x-b)$

$\large f(c)=(c-a)(c-b)=(a-c)(b-c)$

$\large f(-d)=(-d-a)(-d-b)=(a+d)(b+d)$

given expression is equal to $\large f(c)\times f(-d)=(c^2+pc+1)(d^2-pd+1)$

$\large g(x)=x^2+qx+1$

Since $\large c$ and $\large d$ are roots,

$\Rightarrow \large c^2+1=-qc$ ,

$\large d^2+1=-qd$

$\large f(c)\times f(-d)=(pc-qc)(-pd-qd)=-cd(p^2-q^2)$

$\large cd=1$ (product of roots)

$\large f(c)\times f(-d)=-(p^2-q^2)=\boxed{(q^2-p^2)}$

i did it without pen-paper , just observation , that's the real mental prowess :) ..this is a very easy question , it can be used on tougher ones too.. so , instead of thinking about only 1 question , think how you can increase your overall efficiency :) .... if p=q , you should get a 0 so , A or C might be correct but since brilliant shuffles the answers so the probability of it being A are much lower than anyone else , so C done if not this way , it can be done easily algebraically too .