Only One Solution Is Trivial

Relevant wiki: Diophantine Equations - Solve by Factoring

From $x+y+z=3$ we get $x=3-y-z$

Then $3=x^3+y^3+z^3=(3-y-z)^3+y^3+z^3=27-27y+9y^2-27z+18yz-3y^2z+9z^2-3yz^2$

$\implies$ $24=27y-9y^2+27z-9z^2-18yz+3yz^2+3y^2z$

$\implies$ $8=9y-3y^2+9z-3z^2-6yz+yz^2+y^2z$

$\implies$ $8=(3-z)(y+z)(3-y)$ (1)

So $3-z$ is a divisor (positive or negative) of 8

Then $3-z\in$ {1, 2, 4, 8, -1, -2, -4, -8} $\implies$ $z\in$ {11, 7, 5, 4, 2, 1, -1, -5}

Analogously $x\in$ {11, 7, 5, 4, 2, 1, -1, -5} and $y\in$ {11, 7, 5, 4, 2, 1, -1, -5}

We also know from equation 1 that $y+z$ is also a divisor of 8 and analogously $x+z$ and $y+x$ are also divisors of 8. So for any pair of numbers we choose from {11, 7, 5, 4, 2, 1, -1, -5} their sum must be a divisor of 8.

The only triples formed with these numbers that satisfy that any pairwise sum is a divisor of 8 are:

$(7, 1, 1)$ , $(7, -5, 1)$ , $(5, -1, -1)$ , $(4, 4, 4)$ , $(4, 4, -5)$ , $(2, 2, 2)$ , $(2, 2, -1)$ , $(2, -1, -1)$ , $(1, 1, 1)$ , $(1, 1, -5)$ , $(-1, -1, -1)$

From which the only ones that add up to 3 are:

$(7, -5, 1)$ , $(5, -1, -1)$ , $(4, 4, -5)$ , $(2, 2, -1)$ , $(1, 1, 1)$

From which the only ones whose cubes add up to 3 are:

$(4, 4, -5)$ and $(1, 1, 1)$

So the only possible values of $xyz$ are $4\times4\times-5=-80$ and $1\times1\times1=1$ which add up to $\boxed{-79}$

Given:x+y+z=3 x^3+y^3+z^3=3

x^3+y^3+z^3-3xyz=(x+y+z)((x+y+z)^2-3(xy+yz+zx)

xyz-3(xy+yz+zx)=-8

xyz-3(xy+yz+zx)+9(x+y+z)-27=-8

(x-3)(y-3)(z-3)=-8=abc for some integers a,b,c

Let x=a+3,y=b+3,z=c+3

Since x+y+z=3 So a+b+c=-6

Thus (a,b,c)=(-2,-2,-2),(1,1,-8),(-8,1,1),(1,-8,1) and (x,y,z)=(1,1,1),(4,4,-5),(-5,4,4),(4,-5,4)

Sum of distinct values of xyz=(1)+(-80)=-79