Random Problem

It's a sum of two cubes : 27,000,001 = 300³ + 1³

a³ + b³ = (a + b)(a² - ab + b²)

a³ + 1 = (a + 1)(a² - a + 1)

300³ + 1 = 301 (90000 - 300 + 1) = 301 89701

Now, 301 = 7*43 ,So now you have to factor 89701.I know of no way better than just to keep trying successive prime divisors until you find one that works. You have to go only as high as the largest prime whose square is ≤ the number, and that's 293 in this case.

89701 = 271*331 = (301 - 30)(301 + 30) = 301² - 30² = 90601 - 900

So, 27000001 = 7 * 43 * 271 * 331.

Therefore sum of the four prime factors = 4+43+271+331=652.

$27000001=27000000+1=300^3+1^3=(300+1)(300^{2}-300+1)=7 \times 43 \times 89701$ $89701=90601-900=301^2-30^2=331 \times 271$