Question 1104281
For n=2:  
{{{ 1^3 + 2^3 = 9  }}}
and
 {{{ ((2^2*(2+1)^2) / 4) = (4*9)/4 = 9  }}}
—
For n=3:
{{{ 1^3 + 2^3 + 3^3 = 1+8+27 = 36 }}}
and
{{{  ((3^2*(3+1)^2) / 4) = (9*16)/4 = 36 }}}
—
Assume it is true for n=k.   That is  {{{ 1^3+2^3 }}} + ... + {{{ (k-1)^3 + k^3  = (k^2(k+1)^2) / 4 }}}   (*)
—
Now let n=k+1:
 {{{ 1^3 + 2^3 + 3^3 }}} + ... + {{{ k^3 + (k+1)^3 }}}
Which, upon substituting (*), gets us to:
 = {{{ (k^2(k+1)^2)/4 + (k+1)^3 }}} 
factoring:
= {{{ ((k+1)^2)/4 * (k^2 + (4k+ 4)) }}} 
= {{{ ((k+1)^2)/4 * (k+2)^2 }}}

QED.   
(Assuming (*) is true for n=k leads to (*) being true for n=k+1)