2021/035) Let $x,\,y,\,z$ be integers such that $(x-y)^2+(y-z)^2+(z-x)^2=xyz$, prove that $x^3+y^3+z^3$ is divisible by $x+y+z+6$.

 We know $x^3+y^3+z^3 - 3xyz = \frac{1}{2}(x+y+z)((x-y)^2 + (y-z)^2 + (z-x)^2)$

Hence  $x^3+y^3+z^3 = 3xyz + \frac{1}{2}(x+y+z)((x-y)^2 + (y-z)^2 + (z-x)^2)$

Hence $x^3+y^3+z^3 = 3xyz + \frac{1}{2}(x+y+z)(xyz)$ (putting the value from given condition)

Or $x^3+y^3+z^3 = xyz( 3 + \frac{1}{2}(x+y+z))$

Or $x^3+y^3+z^3 = \frac{xyz}{2}( 6 + x+y+z)$

If we can prove that xyz is even then  we are through

As (x-y), (y-z) and (z-x) sum to give zero so atleast one of them is even. So xyz is even from the given condition so  $\frac{xyz}{2}$ is an integer and hence $x^3+y^3+z^3$ is multiple of $(6 + x+y+z)$