In general for $f(x)$ irreducible in $K[x]$ (for $K$ a field) we look for a field extension of $K$ having a root of $f(x)$ as $K[x]/\langle f(x) \rangle $.
In the case of $x^3-x+1$ over $\mathbb{F}_3$, I have found $\mathbb{F}_3[x]/(x^3-x+1)$ to be the splitting field of $x^3-x+1$.
I do not think this would be the case for all such polynomials and all such fields.
Could someone help me to know what is special in this case??
On
you can show this fact for finite fields because all extensions are Galois everywhere! Namely, the splitting field of an irreducible $p$ of degree $n$ will be contained in $F_{p^n}$ since $x^{p^n}-x$ contains all irreds of degree $n$. So the splitting field of $p$ will be a subfield of $F_{p^n}$ which corresponds to a subgroup of $Z/nZ$ which is normal, hence the extension is Galois. Since in a Galois extension having one root means you have all roots, your field is splitting.
Your observation is true, if $K$ is a finite field.. It is also true, if $K=\mathbb{R}$, because the only algebraic extension of that field is algebraically closed. If $\deg f(x)=2$, then it also holds because the sum of the two roots is in $K$, so if you join one, you automatically also join the other.
Over other fields it may or may not hold. For example over $\mathbb{Q}$ the claim does not hold, when $f(x)=x^3-2$, because exactly one of the roots of that polynomial is real, so joining that real root won't give you the rest. On the other hand, if $f(x)=x^3+x^2-2x-1$, then $\mathbb{Q}[x]/(f(x))$ is the splitting field of $f(x)$, because the zeros of $f(x)$ are $2\cos 2\pi/7$, $2\cos4\pi/7$ and $2\cos8\pi/7$. If you join one of those roots, you will also join the others because $$ \cos2\alpha=2\cos^2\alpha-1. $$ I think that, in a sense that I cannot make precise, it is rare for a field $K$ to have this property for all irreducible polynomials $f(x)$.