The number of ordered pairs of integers $(m,n)$ for which $mn \ge 0$ and

$m^3 + n^3 + 99mn = 33^3$
is equal to
$\mathrm{(A) \ }2 \qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 33\qquad \mathrm{(D) \ }35 \qquad \mathrm{(E) \ } 99$
We recall the factorization (see [elementary symmetric sums](https://artofproblemsolving.com/wiki/index.php/Elementary_symmetric_sum))
\[x^3 + y^3 + z^3 - 3xyz = \frac12\cdot(x + y + z)\cdot((x - y)^2 + (y - z)^2 + (z - x)^2)\]
Setting $x = m,y = n,z = - 33$, we have that either $m + n - 33 = 0$ or $m = n = - 33$ (by the [Trivial Inequality](https://artofproblemsolving.com/wiki/index.php/Trivial_Inequality)). Thus, there are $35 \Longrightarrow \mathrm{(D)}$ solutions satisfying $mn \ge 0$.