1±2±3±...±n =(1+n)*n/2

plugging that into the right side of the equation to transform it:

((1+n)*n/2)^2 = (1+n)^2*n^2/4=n^2(n^2+2n+1)/4 = (n^4 + 2n^3 +n^2)/4

If this holds for n: 1^3 + 2^3 +3^3 + ... + n^3 = (n^4 + 2n^3 +n^2)/4

Then for n+1: (n^4 + 2n^3 +n^2)/4 + (n+1)^3 =? (1+n + 1)^2*(n+1)^2/4

(n^4 + 2n^3 +n^2)/4 + (n+1)^3 =? (n^2+4n + 4)(n^2 +2n + 1)/4

(n^4 + 2n^3 +n^2)/4 + (n+1)^3 =? (n^4 + 4n^3 + 4n^2 + 2n^3 + 8n^2 + 8n + n^2 + 4n + 4)/4

(n^4 + 2n^3 +n^2)/4 + (n+1)^3 =? (n^4 + 2n^3 + n^2)/4 + (4n^3 + 12n^2 + 12n + 4)/4

(n+1)(n^2 +2n + 1) =? n^3 + 3n^2 + 3n + 1

n^3 + 2n^2 + n + n^2 + 2n + 1 =? n^3 + 3n^2 + 3n + 1

n^3 + 3n^2 + 3n + 1 =? n^3 + 3n^2 + 3n + 1

Which is obviously true.

So yes, it holds forever.