## Cube expressible as a sum of consecutive cubes in two distinct ways

The sum of   $k$   consecutive cubes beginning with   $m^3$   is :

$m^3 \; + \; (m+1)^3 \; + \; (m+2)^3 \; + \; \dotsb \; + \; (m+k-1)^3$
$= \; (k/4) \; (k + 2 \, m - 1) \, (k^2 + 2 \, k \, m - k + 2 \, m^2 - 2 \, m)$

The cube   $2856^3$   is expressible as a sum of consecutive cubes in two distinct ways:

$2856^3 \; = \; 213^3 \; + \; 214^3 \; + \; 215^3 \; + \; \dotsb \; + \; 555^3$
$2856^3 \; = \; 273^3 \; + \; 274^3 \; + \; 275^3 \; + \; \dotsb \; + \; 560^3$

Can you find another example?