Two symmetric systems

1. Elementary symmetric polynomials. $e_1=3$. From $e_1^2=p_2+2e_2$: $9=5+2e_2$, so $e_2=2$. Newton's identity gives
$$9=3\cdot5-2\cdot3+3e_3=9+3e_3,\quad\text{so } e_3=0.$$
Therefore $x,y,z$ are the roots of $t^3-3t^2+2t=t(t-1)(t-2)$. The solutions are all permutations of $(0,1,2)$.

Check. $0+1+2=3$, $0+1+4=5$, $0+1+8=9$ ✓.

2. Expanding,
$$(x-1)^2+(y-1)^2+(z-1)^2=3-6+3=0.$$
A sum of three real squares is zero only if each is zero: the only solution is $(1,1,1)$.

Lesson. A symmetric system is solved through $e_1,e_2,e_3$ and the cubic whose roots are the unknowns.