Could anyone please solve this question from algebra?

Prove #x^3 + y^3+z^3-3xyz=1/2(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]#

1 Answer
Jun 18, 2018

Please see a Proof in the Explanation.

Explanation:

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!