2026/025) Show that the sum of three consecutive perfect cubes can always be written as the difference between two perfect squares.

We have sum of 1st n cubes ($\frac{(n(n+1)}{2})^2$

Hence $\sum_{k=1}^{k=n} k^3$ =  $(\frac{(n(n+1)}{2})^2$

Also  sun of 1st n+p cubes is

 $\sum_{k=1}^{k=n+p} k^3 $ =  $(\frac{((n+p)(n+p+1)}{2})^2$

Hence sum of p cubes from $(n+1)^3$ to $(n+p)^3$ is

  $\sum_{k=n+1}^{k=n+p} k^3 $ =  $(\frac{((n+p)(n+p+1)}{2})^2$ - $(\frac{(n(n+1)}{2})^2$

p = 3 is a special case of the problem