Visual representation of the surprising validity of this equality...

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.

Yes, assuming it's true, it's true for both m and m-1. Lets substract both-sides of equations for m and m-1 and we get:

m^3 = ( 1+2+3+...m)^2 - (1+2+3...(m-1))^2

Rth side is two arithmetic series. We can simplify:

m^3 = ( (1+m) * (m) / 2 )^2 - ( (1+m-1) * (m-1) / 2) ) ^ 2
m^3 = ( (m+1) * m / 2 )^2 - ( (m-1) * m / 2 )^2

Divide both sides by (m / 2)^2

4m = ( (m+1) )^2 - ( (m-1) )^2
4m = ( (m+1 + m-1) ) * ( (m+1-m+1) )
4m = ( 2m ) * ( 2 )
4m = 4m

For reference, I just posted the video that I promised in the context of our infinity problem from 2 days ago: #705385

My intuition says no, but I don't recall how to demonstrate that.