Question

If $x^3+y^3+z^3=6xyz$, where x, y, z are non-zero integers, $x+y+z\ne 0$ and $xyz=x+y+z$, then the value of $(x–y{)}^2+(y–z{)}^2+(z–x{)}^2$ is equal to

• A. 0
• B. 3
• C. 6
• D. 12

Answer 1

The correct option is C 6

Given that $x^3+y^3+z^3=6xyz$ and $x+y+z=xyz$ such that $x+y+z\ne 0.$
Now, $(x–x+y–y+z–z)=0$
$\Rightarrow \left(\right(x–y)+(y–z)+(z–x\left)\right)=0$
Squaring both sides, we get:
$\left[\right(x–y)+(y–z)+(z–x){]}^2=0$
$\Rightarrow (x-y{)}^2+(y-z{)}^2+(z-x{)}^2+2(x-y\left)\right(y-z)+2(y-z\left)\right(z-x)+2(z-x\left)\right(x-y)=0$
[$\therefore (x+y+z{)}^2=x^2+y^2+z^2+2(xy+yz+zx)$]
$\Rightarrow (x-y{)}^2+(y-z{)}^2+(z-x{)}^2=-2[(x-y)(y-z)+(y-z)(z-x)+(z-x)(x-y)]$
$\Rightarrow (x-y{)}^2+(y-z{)}^2+(z-x{)}^2=-2[xy-y^2-zx+yz+yz-z^2-xy+zx+zx-x^2-yz+xy]$
$\Rightarrow (x-y{)}^2+(y-z{)}^2+(z-x{)}^2=-2[-x^2-y^2-z^2+xy+yz+zx]$
$\Rightarrow (x-y{)}^2+(y-z{)}^2+(z-x{)}^2=-2[x^2+y^2+z^2-xy-yz-zx]$ .......(i)
Consider the identity of $(x^3+y^3+z^3–3xyz)$.
$\therefore x^3+y^3+z^3–3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$
$\Rightarrow 6xyz-3xyz=xyz(x^2+y^2+z^2-xy-yz-zx)$ [$\therefore x^3+y^3+z^3=6xyz$ and $x+y+z=xyz$]
$\Rightarrow 3xyz=xyz(x^2+y^2+z^2-xy-yz-zx)$
$\Rightarrow x^2+y^2+z^2-xy-yz-zx=3$
From (i) and (ii), we get:
$(x–y{)}^2+(y–z{)}^2+(z–x{)}^2=2\times 3=6$
Hence, the correct answer is option (3).