Find GCD!

$p^{2}=27q=3^{2}\cdot3q$

Since they are all integers, $q$ must be a multiple of 3: let $q = 3r$ . $p$ also is a multiple of 3 in that case: let $p = 3t$ .

$9t^{2}=81r$

=> $r=(t/3)^{2}$

As $r$ is an integer, $t$ must be a multiple of 3.

So, let $p = 9a$ and $q=3a^{2}$

We know that,

Product of two numbers = $GCD$ $\times$ $LCM$

$x^{2}y \times xy^{2}= p \times q$

$x^{3}y^{3}=27a^{3}$

$x \times y=3 \times a$

Since $a$ can be a prime number or a composite, in general we can say that: $GCD$ ( $x, y)=3$ and $LCM$ ( $x, y)=a$