1^3+2^3+3^3+................+n^3

1^3+2^3+3^3+................+n^3 = [n(n+1)/2]^2 = n^2(n+1)^2 /4

To prove this, let S(m) be the sum 1^m + 2^m + 3^m +...+ n^m
Consider k^4 - (k-1)^4
k^4 - (k-1)^4
= k^4 - (k^4 - 4k^3 + 6k^2 - 4k +1)
= 4k^3 - 6k^2 + 4k -1

Taking sum over k from 1 to n, we have:
Σ_(k=1 to k=n) (k^4 - (k-1)^4) = Σ_(k=1 to k=n) (4k^3 - 6k^2 + 4k -1)
n^4 - 0^4 = 4S(3) - 6S(2) + 4S(1) - S(0)
n^4 = 4S(3) - 6[n(n+1)(2n+1)/6] + 4[n(n+1)/2] - n
4S(3) = n^4 + 6[n(n+1)(2n+1)/6] - 4[n(n+1)/2] + n
4S(3) = n^4 + n(n+1)(2n+1) - 2n(n+1) + n
4S(3) = n[n^3 + (n+1)(2n+1) - 2(n+1) + 1]
4S(3) = n(n^3 + 2n^2 + n)
4S(3) = n^2 (n^2 + 2n + 1)
4S(3) = n^2 (n+1)^2
S(3) = n^2 (n+1)^2 / 4

1^3+2^3+3^3+................+n^3

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