How many natural number triples $(x,y,z)$ are there such that $xyz = 10^6$?

$ 
\textbf{(A)}\ 568
\qquad\textbf{(B)}\ 784
\qquad\textbf{(C)}\ 812
\qquad\textbf{(D)}\ 816
\qquad\textbf{(E)}\ 824
$
1. We start with the equation \(xyz = 10^6\). We can factorize \(10^6\) as follows:
   \[
   10^6 = (2 \cdot 5)^6 = 2^6 \cdot 5^6
   \]
   Therefore, we need to find the number of natural number triples \((x, y, z)\) such that \(x, y, z\) are products of the factors \(2^a \cdot 5^b\) where \(a + b = 6\).

2. We treat this as a distribution problem where we need to distribute the factors of \(2\) and \(5\) among \(x, y,\) and \(z\). 

3. First, we distribute the factors of \(2\). We have 6 indistinguishable factors of \(2\) to distribute among 3 distinguishable variables \(x, y,\) and \(z\). The number of ways to do this is given by the stars and bars theorem:
   \[
   \binom{6 + 3 - 1}{3 - 1} = \binom{8}{2} = 28
   \]

4. Next, we distribute the factors of \(5\) in the same manner. We have 6 indistinguishable factors of \(5\) to distribute among 3 distinguishable variables \(x, y,\) and \(z\). The number of ways to do this is also:
   \[
   \binom{6 + 3 - 1}{3 - 1} = \binom{8}{2} = 28
   \]

5. Each unique distribution of the factors of \(2\) and \(5\) corresponds to a unique solution for \((x, y, z)\). Therefore, the total number of natural number triples \((x, y, z)\) such that \(xyz = 10^6\) is:
   \[
   28 \times 28 = 784
   \]

The final answer is \(\boxed{784}\)