Clever manipulations 2

There are two key insights to this problem. Firstly, notice that $p+q+r=2$ by Vieta or $q+r=2-p$ . Thus $p^2+q+r=p^2-p+2$ . Doing similarly for the other two, we get the desired product as

$(p^2-p+2)(q^2-q+2)(r^2-r+2)$

The second clever idea is to note that since $p$ is a root, then $p^3-2p^2+2p-4=0$ or $p^3=2p^2-2p+4$ and thus $p^2-p+2=\frac{p^3}{2}$ . Doing similarly for the other two, the product is

$\frac{(p^3)(q^3)(r^3)}{2^3}$

and since $pqr=4$ by Vieta, the answer is

$\frac{4^3}{2^3}=\boxed{8}$

$f(x) = x^{2}(x - 2) + 2(x - 2)$

$f(x)= (x - 2)( x^{2} + 2)$

$f(x) = 0 , x = 2 , x^{2} = -2$

$then p = 2 , q = i \sqrt {2} , r = - i \sqrt{2}$

thus kepping the values we get answer as 8