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- Feb 14, 2012

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- Thread starter anemone
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- Feb 14, 2012

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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 $(6 + x+y+z)$