Completing Cube

The answer is NO that
"There exists a positive integer M,
　such that 1³ + 2³ + … + n³ = M³, where n > 1."

First of all, we know a common result about the sum of cubes:
1³ + 2³ + … + n³ = [ n(n+1)/2 ]²

Therefore, the statement is equivalent to
"There exists a positive integer M,
　such that [ n(n+1)/2 ]² = M³, where n > 1."

If a number is both a perfect square and a perfect cube, it must be a perfect sixth power.

That is, [ n(n+1)/2 ]² = M³ = A⁶ where A is a positive integer.

In other words, we need that
n(n+1)/2 = A³ is a perfect cube, and M = A² is a perfect square.

(Actually, the only triangular number n(n+1)/2 which is also a perfect cube is only 1.)

Consider
n(n+1)/2 = A³
n² + n = 2A³
4n² + 4n = 8A³
4n² + 4n + 1 = 8A³ + 1
(2n + 1)² = (2A)³ + 1
(2n + 1)² - (2A)³ = 1 ...(✿)

Among all integers p and q such that p² - q³ = 1,
the ONLY solution is p = 3 and q = 2.
This is called the Catalan's conjecture (now a theorem).

In other words, the only solution to (✿) is that
2n + 1 = 3 and 2A = 2M² = 2
That is, n = 1 and M = 1.

Therefore, the ONLY possible case is n = 1.

That means, there does not exists a positive integer M
　such that 1³ + 2³ + … + n³ = M³, where n > 1.

References:
http://en.wikipedia.org/wiki/Squared_triangular_number

http://mathworld.wolfram.com/CubicTriangularNumber.html

http://mathworld.wolfram.com/CatalansConjecture.html

Basic references:
http://www.themathpage.com/arith/appendix.htm

http://mathcentral.uregina.ca/QQ/database/QQ.09.06/h/arul1.html