Take a glance at the following sums of cubes.

                                                  1³                   =        1      =     1²       =           (1)²   

                                             1³ + 2³                =        9      =     3²       =        (1 + 2)²

                                          1³ + 2³ + 3³           =      36      =     6²      =      (1 + 2 + 3)²

                                     1³ + 2³ + 3³ + 4³        =    100      =   10²      =   (1 + 2 + 3 + 4)²

         It appears from these four equations that to find the sum of the cubes of the first n consecutive natural

     numbers, all that is necessary is to add up these natural numbers and square the result. For example, to

     find 1³ + 2³ + 3³ + 4³ + 5³ + 6³, simply find the sum 1 + 2 + 3 + 4 + 5 + 6, which is 21, square this to get

     21² = 441, and you have the sum of the cubes of the first six natural numbers without having to calculate

     any cubes at all!      

         In general,                                 1³  +  2³  +  .  .  .  +   n³   =   (1 + 2 +  . . . + n)²                                        (1)

    To prove (1), we use the two formulas:

                         1³ + 2³ + .  .  .  + n³  =  n² (n+1)²/4        and      1 + 2 + .  .  .  + n  =  n(n+1) / 2.

    Then,   1³ + 2³ + . . .  + n³  =  n² (n+1)² / 4  =  [ n(n+1) / 2 ]²  =  (1 + 2 + . . .  + n)², and (1) is proved.

     add up the first eight natural numbers; 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36, and then square the result:

     36²  = 1296.   Thus,   1³  +  2³  +  3³  +  4³  +  5³  +  6³  +  7³ +  8³   =   1296!