If $\gcd(a,b)=9,$ then what is $\gcd(a^2,b^3)\,?$ I know that by the euclidean algorithm, I can obtain the following equations.
I tried some algebraic manipulation but I can't seem to determine, if $$\text{if }\;\gcd(a,b)=9,\text{ then what is }\gcd(a^2,b^3)\;?$$

If $\gcd(a, b) = 9$, that means that $a = 9m$, $b = 9n$, and $\gcd(m, n) = 1$.
Just in case you have any doubt on that last point, let's say $k = \gcd(m, n) > 1$. Then $k$ is a divisor of both $a$ and $b$, and in fact $\gcd(a, b) = 9k > 9$. But if $k = \gcd(m, n) = 1$, that preserves $\gcd(a, b) = 9$.
With that out of the way, we can tackle $\gcd(a^2, b^3)$. We see that $a^2 = 81m^2$ and $b^3 = 729n^3$. Since squaring and cubing add no new distinct prime factors, $\gcd(m^2, n^3) = 1$. This problem then boils down to $\gcd(81, 729) = 81$.
Try plugging in some specific values: $\gcd(18, 45) = 9$. Then $\gcd(18^2, 45^3) = \gcd(324, 91125) = 81$.
