Master theorem and constants independent of $n$
Use a recursion tree (otherwise known as the proof of the Master Theorem).  I'll assume here that $r&gt;1$, since otherwise the recurrence never bottoms out.


The root of the recursion tree for $T(n)$ has value $nr^2$.
The root has $r^2$ children, each with value $(n/r)r^2 = nr$.  Thus, the total value of all children is $nr^3$.
Each child has $r^2$ children of its own, each with value $(n/r^2)r^2 = n$.  Thus, the total value of all grandchildren of the root is $nr^4$.
An easy induction argument implies that for any integer $\ell\ge 0$, the root has $r^{2\ell}$ descendants at level $\ell$, each with value $(n/r^\ell)r^2 = nr^{2-\ell}$.  Thus, the total value of all nodes at level $\ell$ is $r^{2\ell} \cdot nr^{2-\ell} = nr^{2+\ell}$.
The level sums $T(n) = nr^2 + nr^3 + \cdots + nr^{2+\ell} + \cdots$ form an increasing geometric series, so (for big-Oh purposes) only the deepest level matters.
The recursion reaches its base case when $n/r^\ell = 1$, or equivalently, when $\ell = \log_r n$.
The total value of all nodes at that level is $nr^{2+\ell} = nr^{2+\log_r n} = n^2r^2$.
We conclude that $T(n) = O(r^2 n^2)$.


These notes describe the recursion tree method in more detail, with lots of examples.

Equivalently, define $S(n) = T(n)/r^2$, so that $S(n) = r^2 T(n/r) + n$, and then use the Master theorem to solve for $S(n)$.  But that would require remembering the Master theorem, which I don't.