I asked this question here before too, but vaguely, hopefully, this time will be a better attempt:
There are nonzero integers $a$, $b$, $r$, and $s$ such that the complex number $r+si$ is a zero of the polynomial $P(x) = x^3 - ax^2 + bx - 65$. For each possible combination of $a$ and $b$, let $p_{a,b}$ be the sum of the zeroes of $P(x)$. Find the sum of the $p_{a,b}$'s for all possible combinations of $a$ and $b$.
Let the integer zero be $k$
I got that:
$a=2r+k$
$b=2rk+r^2+s^2$
$65=n(r^2+s^2)$
Then I got:
Cases, not in order: $r^2 + s^2 = 13, r^2 + s^2 = 1, r^2 + s^2 = 5, r^2 + s^2 = 65$
For case 1: $r = \{2, -2, 1, -1 \}$
$\sum a = 2(\sum r) + k \implies a = 13$
Then for case 2: $r^2 + s^2 = 13$, it is that,
$\sum a = 2(\sum r) + k \implies a = 5$
Case 3:
$\sum a = 2(\sum r) + k \implies a = 65$
Case 4:
$\sum a = 2(\sum r) + k \implies a = 1$
$$\sum a = 1 + 65 + 5 + 13 = 84$$
The answer is $(80)$, but I got $84$?