FACTORIAL-CUBES !

This is based on the Identity ${ \left( a+b+c \right) }^{ 3 }\quad =\quad { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }-3\left( a+b \right) \left( b+c \right) \left( c+a \right)$ ,In the given equation $\frac { { \left( x+y+z \right) }^{ 3 }-{ \left( x+y-z \right) }^{ 3 }-{ \left( y+z-x \right) }^{ 3 }-{ \left( z+x-y \right) }^{ 3 }-23xyz }{ 2013! }$

Let x+y-z = a, y+z-x = b, z+x-y = c

$\Rightarrow$ a+b+c = x+y+z and a+b = 2y, b+c = 2z, c+a = 2x.

Given expression

$={ \left( a+b+c \right) }^{ 3 }-{ a }^{ 3 }-{ b }^{ 3 }-{ c }^{ 3 }-23xyz\quad /2013!\\ ={ \left( a+b+c \right) }^{ 3 }-{ a }^{ 3 }-{ b }^{ 3 }-{ c }^{ 3 }-24xyz\quad +\quad xyz\quad /2013!\\ ={ \left( a+b+c \right) }^{ 3 }-{ a }^{ 3 }-{ b }^{ 3 }-{ c }^{ 3 }-3\left( a+b \right) \left( b+c \right) \left( c+a \right) \quad +\quad xyz\quad /2013!\\ =0\quad +\quad xyz\quad /2013!\quad =\quad 2014!\times 2015!\times \frac { 1 }{ 2015! } \quad /2013!\\ \Rightarrow \frac { 2014! }{ 2013! } \quad =\quad 2014.$