8. Let the equation $x y z=900$ have all positive integer solutions $\left(x_{i}, y_{i}, z_{i}\right)(1 \leqslant i \leqslant n)$. Then $\sum_{k=1}^{n}\left(x_{k}+y_{k}+z_{k}\right)=$
8. 22572 .

Notice that, $900=2^{2} \times 3^{2} \times 5^{2}$.
Fix $x=2^{\alpha} \times 3^{\beta} \times 5^{\gamma}(0 \leqslant \alpha, \beta, \gamma \leqslant 2)$, at this time,
$$
\begin{array}{l} 
yz=2^{2-\alpha} \times 3^{2-\beta} \times 5^{2-\gamma} . \\
\text { Then } \sum_{k=1}^{n}\left(x_{k}+y_{k}+z_{k}\right) \\
= \sum_{\alpha, \beta, \gamma}\left((3-\alpha)(3-\beta)(3-\gamma) 2^{\alpha} \times 3^{\beta} \times 5^{\gamma}+\right. \\
\left.2 \times \frac{2^{3-\alpha}-1}{2-1} \times \frac{3^{3-\beta}-1}{3-1} \times \frac{5^{3-\gamma}-1}{5-1}\right) \\
= \sum_{\alpha}(3-\alpha) 2^{\alpha} \sum_{\beta}(3-\beta) 3^{\beta} \sum_{\gamma}(3-\gamma) 5^{\gamma}+ \\
\sum_{\alpha}\left(2^{3-\alpha}-1\right) \sum_{\beta}\left(3^{3-\beta}-1\right) \sum_{\gamma} \frac{5^{3-\gamma}-1}{4} \\
=(3+4+4)(3+6+9)(3+10+25)+ \\
(7+3+1)(26+8+2) \frac{(124+24+4)}{4} \\
= 22572 .
\end{array}
$$