Coincidental relation?

Well the sum of the first n numbers is $\frac{n(n+1)}{2}$ and the sum of the first n cubes is $\frac{n^2(n+1)^2}{4}$ which can be found here . So obviously squaring the first results in the second.

Obligatory Python answer:

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total = 0
n = 1
while total != 666:
    total += n
    n += 1
total = 0
for i in range(1,n):
    total += i**3
i = 1
while 666**i != total:
    i += 1
print ("Answer: 666 ^", i)

It doesn't prove why 666^2 is the answer; but if you cube both sides of the original equation, it is easy to see that the cube of the RHS is greater than 1^3 + 2^3.... n^3. Therefore the answer has to be smaller than 666^3 (the cube of the LHS). Only one option is smaller.

Sum of 1+2+3+... = n(n+1)/2 And the sum of 1^3 + 2^3+3^3+... = (n(n+1)/2)^2. So 666^2 is the anwer