Yes, it's true. I suspect that m is supposed to be a natural number greater than or equal to 1.

Assuming it's true for any one value in the sequence, we can prove that it's true for all succeeding values per induction.

To go from m to m+1, we show that both sides grow the same:

\sum{n=1}^{m+1} n^3 = \left( \sum_{n=1}^{m+1} n \right)^2
(m+1)^3 + \sum{n=1}^m n^3 = \left( m+1+ \sum_{n=1}^{m} n \right)^2
(m+1)^3 + \sum{n=1}^m n^3 = (m+1)^2 + 2*(m+1)*\left(\sum_{n=1}^{m} n\right) + \left( \sum_{n=1}^{m} n \right)^2

Use Gauss sum: (m+1)^3 + \sum{n=1}^m n^3 = (m+1)^2 + 2*(m+1)m(m+1)/2 + \left( \sum_{n=1}^{m} n \right)^2

Simplify: (m+1)^3 + \sum{n=1}^m n^3 = (m+1)^2 + m*(m+1)^2 + \left( \sum_{n=1}^{m} n \right)^2
1x+nx=(1+n)x: (m+1)^3 + \sum{n=1}^m n^3 = 1(m+1)^2 + m*(m+1)^2 + \left( \sum_{n=1}^{m} n \right)^2
(m+1)^3 + \sum{n=1}^m n^3 = (1+m)*(m+1)^2 + \left( \sum_{n=1}^{m} n \right)^2 (m+1)^3 + \sum{n=1}^m n^3 = (m+1)^3 + \left( \sum_{n=1}^{m} n \right)^2

Deduct (m+1)^3 from both sides: \sum{n=1}^m n^3 = \left( \sum_{n=1}^m n \right)^2

For m = 1: sum(1^3) = (sum(1))^2

It's true for m=1, and we have shown that if it's true for any m, it's also true for m+1. QED.