Note: the original problem mistakenly listed the last number as 26,640. This has been corrected

First, note that 26460=8*27*5*49 -- if we can show that N is divisible by 8, 27, 5, and 7 then it clearly is divisible by their product as these factors are all coprime.

Any odd number squared is one more than a multiple of 8. (2n+1)^2 = 4n^2+4n+1 = 4n(4n+1) + 1. since either n is even (and hence 4n is divisible by 8 or n is odd and 4n+1 is even, the product is a multiple of 8 , then the +1)

Any even number to the 8th power is certainly a multiple of 8 -- it has at least eight twos in its factorization.

So N is of the form (8a+1) - (8b+1) + 8c = 8(a-b+c) which is a multiple of 8, and therefore also a multiple of 4.

Next, note that for a number x, x^4 is one more than a multiple of 5 when x is not itself a multple of 5. x^8 (x^4)^2 has the same property. That means N is of the form 5a -(5b+1) + (5c+1) = 5(a-b+c) which is a multiple of 5.

Next note that all three "base" numbers in N are multiples of three. That means each term has eight threes (minimum) ,  so the sum is a mulitple of 3^8 and hence also a multiple of 3^3=27.

Finally, 27195 is a multiple of 49. Now consider the second two terms: x^8-y^8 = (x^4-y^4)(x^4+y^4) = (x-y)(x+y)(x^2+y^2)(x^4+y^4). Specifically, the second two terms is divisible by (x-y) = 735 (or -735, take your pick). And 735=49*15 so the sum of the second two terms is also divisible by 49 making N divisible by 49.

Since N is a multiple of 49 (7^2), 5, 27 (3^3) and 4 (2^2), it is divisible by 26460.