<s>[INST] Base conversion between base r^3 to r^2 The problem asks the following: 

If $(\alpha2)_{r^3} = (r3\beta)_{r^2},$ find $\alpha, \beta,$ and r.

First off, I assume $r^2$ > 3 and $r^3 > 2$.
I know that $(r3\beta)_{r^2}$ = (10 _ _ _ _ )$_{r}$
I've attempted to convert to base 10 and set them equal to each other:

$(\alpha2)_{r^3} = \alpha ({r^3})^1 + 2({r^3})^0 $
                    = $ (\alpha {r^3} + 2)_{10}$
$(r3\beta)_{r^2} = r({r^2})^2 + 3({r^2})^1 + \beta ({r^2})^0 = (r({r^4}) + 3{r^2} + \beta)_{10}$

and converting to base r:

$\alpha = Mr^2 + Nr + O$
$ 2 = Xr^2 + Y$
$ r = 1r + 0$
$ 3 = Sr + T$
$ \beta = Vr + U$

But I'm pretty much stuck at this point. Don't see any logical progression from here (if I'm even headed in the right direction).
Any help is greatly appreciated.
 [/INST] Let:
\begin{align*}
\alpha &= Ar^2 + Br + C \\
2 &= Dr^2 + Er + F \\
3 &= Gr + H \\
\beta &= Ir + J
\end{align*}
where each uppercase coefficient is some integer in $\{0, 1, \ldots, r - 1\}$. Then by substituting into our expressions in base $10$, we have that:
$$
(Ar^2 + Br + C)r^3 + (Dr^2 + Er + F) = r^5 + (Gr + H)r^2 + (Ir + J)
$$
Comparing coefficients, we have:
\begin{align*}
\boxed{r^5}:\quad A &= 1 \\
\boxed{r^4}:\quad B &= 0 \\
\boxed{r^3}:\quad C &= G \\
\boxed{r^2}:\quad D &= H \\
\boxed{r^1}:\quad E &= I \\
\boxed{r^0}:\quad F &= J \\
\end{align*}
Now let's assume that $r \geq 2$. Then $2 \geq Dr^2 \geq 4D$, which implies that $D = 0$. Hence, $H = 0$ so that $3 = Gr$. But then since $3$ is prime and $r \neq 1$, we have that $G = 1$ and $r = 3$.
Now since $G = 1$, we know that $C = 1$ so that $\alpha = 1(3)^2 + 0(3) + 1 = 10$.
Likewise, since $r = 3$, we know that $D = E = H = I = 0$ and $F = J = 2$ so that $\beta = 0(3) + 2 = 2$.
</s>