Find the roots of all cubics $f(x)$ given $f(2)=1$ and all roots are integral

Since $f$ is monic and all roots are integral, we have $$f(x)=(x-p)(x-q)(x-r)$$ for some integers $p,q,r$. Then $$(2-p)(2-q)(2-r)=f(2)=1$$ Since $(2-p),(2-q),(2-r)inmathbb{Z}$, there are only few possibilities for $p,q,r$, namely $p,q,rin{1,3}$.

Without loss of generality, let $pleq qleq r$. Product of three integers is $1$ only when either (1) all of them are $1$ or (2) one of them is $1$ and the other two are $-1$. Hence we have the following two cases respectively,

$$p=q=r=1tag{1}$$ $$p=1, q=r=3tag{2}$$

Hence there are only two possible monic cubic polynomials

$$f(x)=(x-1)^3tag{$1'$}$$ $$f(x)=(x-1)(x-3)^2tag{$2'$}$$