jmc

There is exactly one term in the simplified expression for every monomial of the form $x^a{y}^b{z}^c$, where $a,b$, and $c$ are non-negative integers, $a$ is even, and $a+b+c=2006$. There are 1004 even values of $a$ with $0\le a\le 2006$. For each such value, $b$ can assume any of the $2007-a$ integer values between 0 and $2006-a$, inclusive, and the value of $c$ is then uniquely determined as $2006-a-b$. Thus the number of terms in the simplified expression is $$(2007-0)+(2007-2)+\cdots +(2007-2006)=2007+2005+\cdots +1.$$This is the sum of the first 1004 odd positive integers, which is $1004^2=\boxed{1,008,016}.$

$$OR$$The given expression is equal to $$\sum \frac{2006!}{a!b!c!}\left(x^a{y}^b{z}^c+x^a(-y{)}^b(-z{)}^c\right),$$where the sum is taken over all non-negative integers $a,b,$ and $c$ with $a+b+c=2006$. Because the number of non-negative integer solutions of $a+b+c=k$ is $\left(\genfrac{}{}{0ex}{}{k+2}{2}\right)$, the sum is taken over $\left(\genfrac{}{}{0ex}{}{2008}{2}\right)$ terms, but those for which $b$ and $c$ have opposite parity have a sum of zero. If $b$ is odd and $c$ is even, then $a$ is odd, so $a=2A+1,b=2B+1,\text{ and }c=2C$ for some non-negative integers $A,B,\text{ and }C$. Therefore $2A+1+2B+1+2C=2006$, so $A+B+C=1002$. Because the last equation has $\left(\genfrac{}{}{0ex}{}{1004}{2}\right)$ non-negative integer solutions, there are $\left(\genfrac{}{}{0ex}{}{1004}{2}\right)$ terms for which $b$ is odd and $c$ is even. The number of terms for which $b$ is even and $c$ is odd is the same. Thus the number of terms in the simplified expression is $$\left(\genfrac{}{}{0ex}{}{2008}{2}\right)-2\left(\genfrac{}{}{0ex}{}{1004}{2}\right)=1004\cdot 2007-1004\cdot 1003=1004^2=\boxed{1,008,016}.$$