How many roots?

From ${ x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 }=x+y+z=3$ and ${ \left( x+y+z \right) }^{ 3 }={ x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 }+3\left( x+y \right) \left( y+z \right) \left( z+x \right)$

we obtain $\left( x+y \right) \left( y+z \right) \left( z+x \right) =8$ or $\left( 3-z \right) \left( 3-x \right) \left( 3-y \right) =8$

On the other hand, $\left( 3-z \right) +\left( 3-x \right) +\left( 3-y \right) -3\left( x+y+z \right) =6$ , 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 $\left| 3-x \right| =\left| 3-y \right| =\left| 3-z \right| =2$ , yielding $x,y,z\in \left\{ 1,5 \right\}$ .

Because $x+y+z=3$ , the only possibility is $x=y=z=1$ .

In the second case, one of $\left| 3-x \right| ,\left| 3-y \right| ,\left| 3-z \right|$ must be 8, say $\left| 3-x \right| =8$ , yielding $x\in \left\{ -5,11 \right\}$ and $\left| 3-y \right| =\left| 3-z \right| =1$ .

Taking into account that $x+y+z=3$ , the only possibility is $x=-5$ and $y=z=4$ .

In conclusion, $\boxed { 4 }$ desired triples are $(1,1,1), (-5,4,4), (4,-5,4)$ and $(4,4,-5)$ .