Cube Sequences

1³=1

1³+2³=9  = 3²                         

1³+2³+3³=36 = 6²

1³+2³+3³+4³=100 = 10²

1³+2³+3³+4³+5³=225 = 15²

…. 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16….

These are triangular numbers are in bold: 

• 1+2 =3
• 1+2+3=6
• 1+2+3+4=10

          ( n(n+1)/2 )²= n²(n+1)²/4

A question I have in my mind is that Fermat’s last theorem states: If an integer n is greater than 2, then the equation aⁿ + bⁿ = cⁿ has no solutions in non-zero integers a, b, and c. 

But how about  a³ + b³ + c³ = d³        … Are there any integer solutions to this?   I ask this because geometrically speaking volume is 1 degree of freedom more than area.

      3³+4³+5³=6³             …. = 15² – 3²