1. Find all integer solutions of the system of equations

$\left\{\begin{array}{c}x+y+z=2, \\ x^{3}+y^{3}+z^{3}=-10 .\end{array}\right.$
Solution:

$\left\{\begin{array}{c}x+y+z=2, \\ x^{3}+y^{3}+z^{3}=-10 .\end{array} \Leftrightarrow \quad\left\{\begin{array}{c}x+y+z=2, \\ (x+y+z)^{3}-x^{3}-y^{3}-z^{3}=18 .\end{array}\right.\right.$

$(x+y+z)^{3}-x^{3}-y^{3}-z^{3}=18 \quad \Leftrightarrow\left((x+y+z)^{3}-x^{3}\right)-\left(y^{3}+z^{3}\right)=18 \Leftrightarrow \Leftrightarrow$

$(y+z)\left((x+y+z)^{2}+x(x+y+z)+x^{2}\right)-(y+z)\left(y^{2}-y z+z^{2}\right)=18 \Leftrightarrow$

$(y+z)\left((x+y+z)^{2}+x(x+y+z)+x^{2}-y^{2}+y z-z^{2}\right)=18 \Leftrightarrow$

$(y+z)((x+y)(x+y+2 z)+x(x+y)+(x+y) z+(x+y)(x-y))=18 \Leftrightarrow$

$(y+z)(x+y)(x+y+2 z+x+z+x-y)=18 \quad \Leftrightarrow(y+z)(x+y)(x+z)=6$.

$y+z=2-x \Rightarrow 4-x$ can take values $\pm 1, \pm 2, \pm 3, \pm 6$

1) $2-x=1, x=1, y+z=1 \Rightarrow z=1-y \Rightarrow(1+y)(2-y)=6 \Rightarrow y^{2}-y+4=0$

No integer solutions.

2) $2-x=-1, x=3, y+z=-1 \Rightarrow z=-1-y \Rightarrow(3+y)(y-2)=6 \Rightarrow y^{2}+y-12=0$

$y=3$ or $y=-4$, we have two solutions of the system with integer values

$x=3, y=3, z=-4$ and $x=3, y=-4, z=3$.

3) $2-x=2, x=0, y+z=2 \Rightarrow z=2-y \Rightarrow(0+y)(2-y)=3 \Rightarrow y^{2}-2 y+3=0$.

No integer solutions.

4) $2-x=-2, x=4, y+z=-2 \Rightarrow z=-2-y \Rightarrow(4+y)(2-y)=-3 \Rightarrow y^{2}+2 y-11=0$.

No integer solutions.

5) $2-x=3, x=-1, y+z=3 \Rightarrow z=3-y \Rightarrow(-1+y)(2-y)=2 \Rightarrow y^{2}-3 y+4=0$.

No integer solutions.

6) $2-x=-3, x=5, y+z=-3 \Rightarrow z=-3-y \Rightarrow(5+y)(2-y)=-2 \Rightarrow y^{2}+3 y-12=0$.

No integer solutions.

7) $2-x=6, x=-4, y+z=6 \Rightarrow z=6-y \Rightarrow(-4+y)(2-y)=1 \Rightarrow y^{2}-6 y+9=0 \Rightarrow y=3$.

We have an integer solution $x=-4, y=3, z=3$.

8) $2-x=-6, x=8, y+z=-6 \Rightarrow z=-6-y \Rightarrow(8+y)(2-y)=-1 \Rightarrow y^{2}+6 y-17=0$.

No integer solutions.

Answer: $(3,3,-4),(3,-4,3),(-4,3,3)$.