We can use the difference of squares to solve this easily.

Note that for any integer $n$, we have:

$(a^{n} - b^{n}) = (a^{n/2} - b^{n/2})(a^{n/2} + b^{n/2})$

So we have:

$(3^{8} - 2^{8})$

$(3^{4} - 2^{4})(3^{4} + 2^{4})$

$(3^{2} - 2^{2})(3^{2} + 2^{2})(3^{4} + 2^{4})$

$(3 - 2)(3 + 2)(3^{2} + 2^{2})(3^{4} + 2^{4})$

$(1)(5)(9+4)(81+16)$

$(1)(5)(13)(97)$

Since 97 is prime, we can no longer factor this expression.

Now we need to multiply the different combinations of these factors to get the complete list of factors for $(3^{8} - 2^{8})$.

Think of factoring 12. This would be $2^{2} * 3$, or $4*3$. Summing 4 and 3 isn't enough, we're forgetting the 1,2,6 and 12.

Permuting the different factors then, we get:

$1,5,13,97,5*13,5*97, 13*97, 5*13*97$

Which simplifies too (I used a calculator here, definitely saves time):

$1,5,13,97,65, 485, 1261, 6305$

Summing these we get $8232$.

Therefore the answer is C