Power of Vieta

Relevant wiki: Vieta's Formula Problem Solving - Intermediate

Since $p$ , $q$ and $r$ are the roots of $f(x) = x^3-19x^2+31x+57$ , this means that $f(x) = (x-p)(x-q)(x-r)$ . And by Vieta's formula we have $p+q+r = 19$ . Then,

$\begin{aligned} (p+q)(q+r)(p+r) & = (19-r)(19-p)(19-q) = f(19) = \boxed{646} \end{aligned}$

Attempting to find the roots of this cubic will most likely prove to be futile, and ugly and not very useful at best.

We begin by applying Vieta's formula to represent the roots in terms of our polynomial's coefficients. It's okay if you're unfamiliar with them, I will derive these relationships here: $(x-p)(x-q)(x-r) = x^3 + ax^2 + bx + c$ $(x-p)(x-q)(x-r) = x^3 - (p+q+r)x^2 + (pq+qr+pr)x - pqr$ $a = - (p+q+r)$ $b = (pq+qr+pr)$ $c = - pqr$

Once we have these formulas, we can manipulate them in order to obtain that $(p+q)(q+r)(p+r) = (p+q+r)(pq+qr+pr) - pqr = \boxed{ c-ab }$

Now we simply plug in our coefficients to get the answer of $57 - (-19)(31) = \boxed{646}$ .

Additional note: Although you are given that the roots of this cubic are all real, it turns out that this formula works perfectly fine with complex values too!