<s>[INST] 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 $x=y=z=1$.
 [/INST] From the identity
$$(x + y + z)^3 = x^3 + y^3 + z^3 + 3(x + y)(y + z)(z + x)$$
we obtain $8 = (x+y)(y+z)(z+x)$. It follows that $(3−x)(3−y)(3−
z) = 8$. On the other hand, 
$$(3−x)+(3−y)+(3−z)−3(x+y+z) = 6,$$
implying that either $|3−x|, |3−y|, |3−z|$ are all even, or exactly one of
them is even. In the first case, we get $|3 −x| = |3−y| = |3−z| = 2$,
yielding $x, y, z \in \{1, 5\}$. Because $x+y+z = 3$, the only possibility is
$x = y = z = 1$. 
In the second case, one of $|3−x|, |3−y|, |3−z|$ must be $8$, say $|3−x| = 8$, yielding $x \in \{−5, 11\}$ and $|3−y| = |3−z| = 1$, from which $y, z \in \{2, z\}$. Taking into account that $x+y +z = 3$, the only possibility is $x = −5$ and $y = z = 4$. In conclusion, the desired
triples are 
$$(1, 1, 1), (−5, 4, 4), (4,−5, 4), \text{ and }(4, 4,−5).$$
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