$\underline{\textbf{Classical Solution}}$ Let $x$, $y$, and $z$ be the roots of the cubic polynomial $$t^3+at^2+bt+c=0$$ Meanwhile, let $$ \left\{ \begin{array}{rcl} S_1=&x+y+z&=3\\ S_2=&x^2+y^2+z^2&=3\\ S_3=&x^3+y^3+z^3&=3 \end{array} \right. $$ Then, by Vieta's theorem, we have $$S_1 = (x+y+z) =-a \implies a=-S_1= {-3}$$ also $$S_2 = x^2+y^2+z^2=(x+y+z)^2 - 2(xy+yz+zx)=a^2 - 2b$$ $$\therefore b=\frac{1}{2}\cdot(a^2 - S_2)={3}$$ and \begin{align*} S_3 &= x^3+y^3+z^3\\ &=(x+y+z)^3 - 3(x+y+z)(xy+yz+zx) + 3xyz\\ &= (-a)^3 - 3(-a)b- 3c \end{align*} $$\therefore\ c =\frac{1}{3}\cdot(-a^3 +3ab - S_3)={-1}$$ Thus $x$, $y$, and $z$ are the roots of the polynomial $$t^3-3t^2+3t-1=(t-1)^3$$ which means $$\boxed{x=y=z=1}$$ $\underline{\textbf{A Quicker Solution}}$ Let $x$, $y$, and $z$ be the three roots of $$P(t)=t^3-at^2+bt-c$$ Then $$0=P(x)+P(y)+P(z)=3-3a+3b-3c\implies 1-a+b-c=0$$ Therefore $$P(1)=1^3-a\cdot 1^2+b\cdot 1 -c =0$$ This means that at least one of $x$, $y$, and $z$ is equal to $1$. Then, by symmetry, we find all of them equal $1$. Hence, there is only one solution $$(x, y, z)=\boxed{(1,1,1)}$$