Solving $x+y+z=x^3+y^3+z^3=3$.

Question

Let $x,y,z$ are non zero integers satisfying the system of equations $x+y+z=3$ and $x^3+y^3+z^3=3$ then find solution triplets (all) of $(x,y,z)$.

I tried using: $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$ but didn't concluded something good, please help to begin with this, in write direction.

Answer 1

The hint: $$27=(x+y+z)^3=x^3+y^3+z^3+3(x+y)(x+z)(y+z)=3+3(x+y)(x+z)(y+z),$$ which gives $$(x+y)(x+z)(y+z)=8$$ and number of cases: $$x+y\in\{\pm1,\pm2\pm4\pm8\}$$

Answer 2

$x+y+z=x^3+y^3+z^3=3$

The solution set for $(x+y)$ in above equation given by Michael Rozenberg is shown below

$(x+y)\in\{\pm1,\pm2\pm4\pm8\}$

The above solution implies that $(x,y,z)$ can only have the below mentioned integer solutions;

$(x,y,z)= (4,4,-5),(-5,4,4),(4,-5,4),(1,1,1)$