Because such roots to polynomial equations come in conjugate pairs, we seek the values $r$ such that $P(x)=r$ has just one real root $x$. Considering the shape of a cubic, we are interested in the boundary values $r$ such that $P(x)-r$ has a repeated zero. Thus, we write $P(x)-r=x^{3}+x^{2}-x+(2-r)=(x-p)^{2}(x-q)=x^{3}-(2 p+q) x^{2}+p(p+2 q) x-p^{2} q$. Then $q=-2 p-1$ and $1=p(p+2 q)=p(-3 p-2)$ so that $p=1 / 3$ or $p=-1$. It follows that the graph of $P(x)$ is horizontal at $x=1 / 3$ (a maximum) and $x=-1$ (a minimum), so the desired values $r$ are $r>P(-1)=3$ and $r<P(1 / 3)=1 / 27+1 / 9-1 / 3+2=49 / 27$.