smc

By the Multinomial Theorem, the summands can be written as $$\sum _{a+b+c=2006}\frac{2006!}{a!b!c!}{x}^a{y}^b{z}^c$$ and $$\sum _{a+b+c=2006}\frac{2006!}{a!b!c!}{x}^a(-y{)}^b(-z{)}^c,$$ respectively. Since the coefficients of like terms are the same in each expression, each like term either cancel one another out or the coefficient doubles. In each expansion there are: $$\left(\genfrac{}{}{0ex}{}{2006+2}{2}\right)=2015028$$ terms without cancellation. For any term in the second expansion to be negative, the parity of the exponents of $y$ and $z$ must be opposite. Now we find a pattern: if the exponent of $y$ is $1$, the exponent of $z$ can be all even integers up to $2004$, so there are $1003$ terms. if the exponent of $y$ is $3$, the exponent of $z$ can go up to $2002$, so there are $1002$ terms. $⋮$ if the exponent of $y$ is $2005$, then $z$ can only be 0, so there is $1$ term. If we add them up, we get $\frac{1003\cdot 1004}{2}$ terms. However, we can switch the exponents of $y$ and $z$ and these terms will still have a negative sign. So there are a total of $1003\cdot 1004$ negative terms. By subtracting this number from 2015028, we obtain $\boxed{1,008,016}$ or $1,008,016$ as our answer.