NMTC 2015

$27000001 = { 300 }^{ 3 }+{ 1 }^{ 3 } \\\ Now\ using\ the\ identity\ { a }^{ 3 }+{ b }^{ 3 }=\left( a+b \right) \left( { a }^{ 2 }+{ b }^{ 2 }-ab \right) \\ We\ get\ 27000001= \left( 301 \right) \left( { 300 }^{ 2 }+{ 1 }^{ 2 }-300 \right) \\ =\left( 301 \right) \left( \left( { 300 }^{ 2 }+{ 1 }^{ 2 }+600 \right) -900 \right) \\ =\left( 7*43 \right) \left( \left( { 300 }^{ 2 }+{ 1 }^{ 2 }+2*1*300 \right) -900 \right) \\ =\left( 7*43 \right) \left( { \left( 301 \right) }^{ 2 }-{ \left( 30 \right) }^{ 2 } \right) \\ =\left( 7*43 \right) \left( \left( 301+30 \right) \left( 300-30 \right) \right) \\ =7*43*331*271 \\ thus\ the\ sum\ of\ the\ four\ prime\ factors\ = 7+43+331+271 = 652$

27000001 = 300^3+1^3 a^3+b^3 = (a+b)(a^2-ab+b^2)

300^3+1=(301)(300^2-300+1) 300^3+1=(301)*(89701)

301 has prime factors of 7 and 43. You know this is divisible by 7 because a number of the form 10x + y is divisible by 7 if and only if x − 2y is divisible by 7 and 30-2(1) = 28. Which is divisible by 7.

89701 is the toughie. You have to do trial by error, I think. Since we obtained the first 2 prime numbers, we know 89701 is a product of 2 primes. So we try prime numbers up until 299. We find that 271 works.

7, 43, 271, 331 = 652