Sum of the factors of a large number

We recognize 27,000,001 as $27\cdot 10^6 + 1$ . This screams sum of cubes. To shorten the writing, let $x=10$ . Now we simply want to factor $27x^6+1$ :

$27x^6 + 1 = (3x^2)^3+1^3 = (3x^2+1)(9x^4-3x^2+1)$

The first term, $3x^2+1$ is order $10^2$ , so we do not algebraically simplify this further; it is small enough to be manually factored. But $9x^4-3x^2+1$ is order $10^4$ , which is not as easily factored manually. We must factor this further algebraically.

Note that $9x^4-3x^2+1$ can be expressed as a difference of squares like so:

$9x^4-3x^2+1 = 9x^4 +6x^2 + 1 - 9x^2$ $=(3x^2+1)^2 - (3x)^2$ $=(3x^2-3x+1)(3x^2+3x+1)$

Both of these terms are now also order $10^2$ and can be easily factored.

Our final product is $27x^6+1 = (3x^2+1)(3x^2-3x+1)(3x^2+3x+1).$ Plugging in $x=10$ , we see that $27,000,001 = 301\cdot 271 \cdot 331.$ Simplifying the product, we see that the prime factorization is $7\cdot 43 \cdot 271 \cdot 331.$ Thus, the sum of the prime factors is $7+43+271+331 = \boxed{652}$ .