Find all Integral solutions to $x+y+z=3$, $x^3+y^3+z^3=3$. Suppose that $x^3+y^3+z^3=3$ and $x+y+z=3$. 
What are all integral solutions of this equation? 
I can only find $x=y=z=1$.

Using identity: $(x+y+z)^3 = x^3+y^3+z^3 + 3(x+y)(y+z)(z+x)$, we have:
$3^3 = 3 + 3(x+y)(y+z)(z+x) \Rightarrow (x+y)(y+z)(z+x) = 8$. From this you should be able to deduce the answer.
