Find the number of positive divisors of $(2008^3 + (3 * 2008 * 2009) + 1)^2$

Question

Find the number of positive divisors of $(2008^3 + (3 * 2008 * 2009) + 1)^2$

What I Tried: I have no idea of an elegant solution to this. I cannot seem to guess or figure out the factors as there is a $(+1)$ added to it, which seems to bother me. As the number is a perfect square I can guess it will have even number of positive divisors or so.

I decided to solve it by hand. I calculated the expression to be $(8108486729)^2$ , and turns out , the prime factorization of $8108486729$ is amazingly $7^6 * 41^3$ .

My first question is, how is that factorization coming? It looks a bit like magic to me, can someone explain that?

So from here $(8108486729)^2$ will have prime factorization as $7^{12} * 41^6$ , so total number of factors will be $(13)(7) = 91$ , problem solved. This is however, not an elegant solution, and I can't seem to find any.

Can anyone help?

Answer 1

Hint: $${(2008)}^3+3\cdot 2008\cdot 2009+1={(2008+1)}^3={(2009)}^3$$

Answer 2

$z^3+3z^2+3z+1=(z+1)^3$

$z=2008$

$z+1=2009$

$2009 = 7^2 * 41$

$2009^3 = 7^6 * 41^3$