Question

If $x^3+y^3+z^3=3xyz$ then the relation between $x,y$ and $z$ is :

• A. $x+y+z=0$
• B. $x=y=z$
• C. Either $x+y+z=0$ or $x=y=z$
• D. $x+y+z=0$ and $x\ne y\ne z$

Answer 1

Answer: (C)

Either $x+y+z=0$ or $x=y=z$

We have $x^3+y^3+z^3-3xyz$

$=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

$0=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

So either $x+y+z=0$ (or)

$x^2+y^2+z^2-xy-yz-zx=0$

$x^2+y^2+z^2-xy-yz-zx=0$

$2x^2+2y^2+2z^2-2xy-2yz-2zx=0$

$(x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)=0$

$(x-y{)}^2+(y-z{)}^2+(z-x{)}^2=0$

$\Rightarrow x=y=z$

Therefore, either $x+y+z=0$ or $x=y=z$.