**Hint.** You need to calculate how many different terms of form $a^nb^mc^k$ can arise from expanding the power. Next you need to figure out what terms do cancel each other (one from $(a + b + c)^{2006}$ and the other from $(a - b- c)^{2006}$) and how many are those.
**Edit**. That hint was not very helpful. So, actually we need to count how many possible nonegative integer values $m, n, k$ satisfy $$ m + n + k = 2006\\\ (-1)^m(-1)^k = 1 \Leftrightarrow m + k \text{ is even} $$ Since $m + k + n = 2006$, $m+k$ is even iff $n$ is even, i.e. $n = 2l$. Finally, the problem is to count how many solutions does the following equation have $$ 2l + m + k = 2006, \qquad l,m,k \geq 0 $$ For a fixed $l$ that equation has $2007 - 2l$ solutions. So the total number of solutions will be $$ \sum_{l=0}^{1003} 2007 - 2l = 1004 \cdot 2007 - 2 \frac{1003 \cdot 1004}{2} = 1004^2 $$