2022/001) Solve in integers $x+y+z=3$ and $x^3+y^3 + z^3 = 3$

We are given

$x+y+ z = 3\cdots(1)$

$x^3 + y^3 + z^ 3 = 3\cdots(2)$

cubing (1) we get $(x+y+z)^3 = 3^3$

or $x^3 + y^3 + z^3 + 3(x+y)(y+z)(z+x) = 27$

or $3 + 3(x+y)(y+z) (z+x) = 27$ putting thw vaue of $x^3 + y^3 + z^3 = 3$ from (2)

or  $(x+y)(y+z)(z+ x) = 8$

now $((x+y) + (y+z) + (z+ x)) = 2 ( x + y + z) = 6$ usng (1)

we need to find 3 numbers whose sum is 6 and product is 8. They are (2,2,2) or (4,1,1) and other combinations fail

taking x + y = 2 , y + z = 2, z+ x = 2 and using (2) that is x + y + z = 3 we get x=1,y=1,z = 1

 taking x + y = 4 , y + z = 1, z+ x = 1 and using (2) that is x + y + z = 3 we get x=5,y=z = - 4

takinn n rotaton other combinations we get x=y = -4, z = 5 as one solution and z = z = -4 and y =5 as another solution

so $\{x,y,z\} = \{2,2,2\}$ or $\{5,-4,-4\}$ (that is any permuation of the same)