Which of the following numbers is divisible by $$3^{10} + 2$$

The given expression, $$3^{10}+2$$, when cubed, gives us $$(3^{10}+2)^3= 3^{30} + 8 + (6\times 3^{20}) + (12\times 3^{10})$$

This expansion can be rewritten as $$(\underline{3^{10}+2)^3} = (3^{30}+8) + \underline{(6\times 3^{10})(3^{10}+2)}$$

Since the two underlined parts are both divisible by $$(3^{10}+2)$$, we can conclude that the remaining non-underlined part, $$(3^{30}+8)$$ should also be divisible by $$(3^{10}+2)$$. Therefore option D, is correct. We can do the same for other options.

We can try $$(3^{10}+2)^2$$ to verify option A, the expansion will be $$(3^{20}+4+4*3^{10})$$. Thus, after dividing this by $$(3^{10}+2)$$ the remainder will clearly be $$4*3^{10}$$, since this is definitely not divisible by $$(3^{10}+2)$$, as $$3^{10}$$ is not divisible by $$(3^{10}+2)$$, which is an odd number, we can conclude option A is incorrect.

Option B is also incorrect because if $$3^{30}+8$$ is divisible after the cubic expansion, a number $$6$$ less than it cannot be.

Finally, if we assume option C to be divisible, then the difference between option C and option D should also be divisible by $$(3^{10}+2)$$, this difference is nothing but $$(3^{30}-3^{20}) = 3^{20}(3^{10}-1)$$, since none of the two parts in this difference are divisible, the difference itself is not divisible, and option C will not be divisible either.

Alternate Explanation:

Let $$3^{10}=a$$ and $$2=b$$

Then the given expression is of form $$a+b$$

So, option A is of form $$a^2+b^2$$

option B is of form $$a^3+b$$

option C is of form $$a^2+b^3$$

option D is of form $$a^3+b^3$$

Among all these polynomials, we know only $$a^3+b^3$$ has a factor $$a+b$$

So, option D is the correct answer.