Prerequisites : # (1) : a^3+b^3=(a+b)^3-3ab(a+b)............(ast)#.
# (2) : c^3+d^3=(c+d)(c^2-cd+d^2)..........................(astast)#.
#"The Expression"=ul(x^3+y^3)+z^3-3xyz#,
#=ul{(x+y)^3-3xy(x+y)}+z^3-3xyz............[because, (ast)]#,
#=(x+y)^3+z^3-ul(3xy(x+y)-3xyx)#,
#=ul(u^3+z^3)-ul(3xyu-3xyz)," say, where, "u=(x+y)#,
#=ul((u+z))(u^2-uz+z^2)-3xyul((u+z))......[because, (astast)]#,
#=(u+z)(u^2+z^2-uz-3xy)#.
Now reverting from #u" to "(x+y)#, we get,
#"The Exp."={(x+y)+z}{(x+y)^2+z^2-(x+y)z-3xy}#,
#=(x+y+z){(x^2+ul(2xy)+y^2)+z^2-ul(3xy)-(x+y)z}#,
#=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)#,
#=1/2(x+y+z)(2x^2+2y^2+2z^2-2xy-2yz-2zx)#,
#=1/2(x+y+z){(x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)},#
#=1/2(x+y+z){(x-y)^2+(y-z)^2+(z-x)^2}#, as desired!
