I am trying to understand a problem in Hungerford (p. 134, Exercise 14). Specifically, it says that if $P$ is a prime ideal in a not necessarily commutative ring $R$, and $r,s\in R$ such that $rRs\subseteq P$, then $r\in P$ or $s\in P$.
A hint is given: suppose that $RrR\subseteq P$, then $(r)^3\subseteq RrR\subseteq P$, and so $r\in (r)\subseteq P$.
But why is $(r)^3\subseteq RrR$? I understand that $(r)$ contains $r$, since by definition it is the smallest ideal to do so. If $(r)^3\subseteq RrR$ then $RrR$ contains $r$. But since $RrR$ is the collection of finite sums of elements of the form $xry$, where $x,y\in R$, I do not see why $RrR$ would contain $r$, in particular if $R$ is without a multiplicative identity.
Note that rings are not necessarily unital in Hungerford, so the ideal $(r)$ consists of elements of the form $$nr + ar + rb + \sum_{i=1}^m a_irb_i\tag{A}$$ where $n\in\mathbb{Z}$, $m$ is a nonnegative integer, and $a,b,a_i,b_i\in R$.
Now, the thing is, the hint does not assert that $RrR$ contains $r$; the assertion is that it contains $(r)^3$. Your claim that if $(r)^3\subseteq RrR$ then $RrR$ contains $r$ is unwarranted. It’s not even true in $\mathbb{Z}$ that $(r)^3$ contains $r$: $(2)^3=(8)$ does not contain $2$.
In order to contain $(r)^3$, given that $RrR$ is certainly closed under sums, it is enough that it contain any triple product of elements as above. And for that, it is enough that it contain any triple product of summands of the element above. This can be checked directly. For example, $$(n_1r)(n_2r)(n_3r) = n_2\Bigl( (n_1r)r(n_3r)\Bigr)\in RrR.$$ Similarly, $$(ar)(rb)(nr) = (ar)r(b(nr))\in RrR$$ and so on. Each term you get when you multiply out three elements of the form (A) is an element of $RrR$, and hence these products all lie in $RrR$, and hence the sums of products of that form lie in $RrR$.