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are 
and 
What is the value of ![\[(p^2 + 4)(q^2 + 4)(r^2 + 4)?\]](http://latex.artofproblemsolving.com/9/3/8/938ecf6869b053c8335864493848733a3f242e2e.png)
to get 
, note that this is the norm of a Gaussian integer, and for parity reasons it should be odd which gives 125
, 
and 
by subbing 
and squaring.
is a root of the given polynomial, then
So the expression becomes ![\[ [(p+1)(q+1)(r+1)]^2 \cdot (1-p)(1-q)(1-r). \]](http://latex.artofproblemsolving.com/0/1/b/01bc78e5244ab321be25f46ffd7902b163e423eb.png)
First part expands to ![\[ pqr + (pq + qr + rp) + (p + q + r) + 1 = -3 - 2 - 1 + 1 = -5 \]](http://latex.artofproblemsolving.com/4/0/4/4047843d99fb60afe3a99afe36f8554838af61ed.png)
and the second part expands to ![\[ -pqr + (pq + qr + rp) - (p + q + r) + 1 = 3 - 1 + 2 + 1 = 5, \]](http://latex.artofproblemsolving.com/1/e/2/1e2ef0a4ba354115a4a0673d8950713ab6470c2c.png)
so substituting yields ![\[ (-5)^2 \cdot 5 = \boxed{125} . \]](http://latex.artofproblemsolving.com/6/d/e/6dea3a9f19bb8734310750d312311da42b832c2f.png)
, as possible:![\[
\frac{1}{pqr}pqr(p^2+4)(q^2+4)(r^2+4)=\frac{1}{pqr}(p^3+4p)(q^3+4q)(r^3+4r)=\frac{-1}{pqr}(2p^2-5p+3)(2q^2-5q+3)(2r^2-5r+3)
\]](http://latex.artofproblemsolving.com/1/7/a/17ac9bad02ff3cf49a8351970a0268f3f6d4ebdc.png)
Which factors nicely into linear products of the roots:![\[
\frac{-1}{pqr}(2p-3)(p-1)(2q-3)(q-1)(2r-3)(r-1)=\frac{1}{3}(-f(1))(-8f(3/2))=\frac{5}{3}\cdot8(6+27/8)=\boxed{125}
\]](http://latex.artofproblemsolving.com/f/7/9/f79e00fee0cae17bb742b2bafca2eb1f6ae34352.png)
Where we have divided out by 
(Vietas). I'm new to Aops but this was too fun a problem to not share!
125?
is the given polynomial, then 
. 
can be found similarly (or just note that only the imaginary part changes), and 
.
.
for all values of . Therefore, the desired expression equals 
yields![\[(p^2+ 4)(q^2+ 4)(r^2 + 4) = (p-2i)(p+2i)(q-2i)(q+2i)(r-2i)(r+2i) = f(2i) f(-2i) = \boxed{125},\]](http://latex.artofproblemsolving.com/0/b/8/0b8e65573369400275dbbac62577ae7a897b2f58.png)
we're done.
but didn't want to do it: I just did 
