Question

$x$ and $y$ are two positive integers. The GCD and LCM of $x^{2}y$ and $xy^{2}$ are $p$ and $q$ respectively. If $p^{2}=27q$, then what is the GCD of x and y?

Answer 1

      p² = 27 q = 3² (3q) 
Since they are all integers,  q must be a multiple of 3.  let q = 3r.  Also p is a multiple of 3.  Let p = 3 t.
      9 t² = 81 r    
=>  r = (t/3)²       
   As r is an integer, t must be a multiple of 3.
        
so  let   p = 9*a   and  q = 3 a²

product of numbers = GCD * LCM
=>    x² y * xy² = p * q
=>    x³ y³   = 27 a³
=>    x y  = 3 * 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
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Another Solution:

Without loss of generality let    e >= g  and   f <= h.  Let  c and d have no factors in common between them.   a and b are prime numbers.

$Let\ x=a^e\ b^f\ c,\ \ \ \ y=a^g\ b^h\ d,\\\\ GCD(x,y)=a^g b^f,\ \ \ \ \ LCM(x,y)=a^e\ b^h \ cd\\\\x^2y=a^{2e+g}\ b^{2f+h}\ c^2\ d,\ \ \ xy^2=a^{e+2g}\ b^{f+2h}\ c\ d^2\\\\As\ e\ \textgreater \ g,\ 2e+g\ \textgreater \ e+2g.\\\\p=a^{e+2g}\ b^{2f+h}\ c\ d,\ \ \ \ q=a^{2e+g}\ b^{f+2h}\ c^2d^2\\\\do\ p^2=27q\ and\ compare\ powers.\\ \implies a^{3g}\ b^{3f}=3^3\ cd\\=\ \textgreater \ a^gb^f=3* \sqrt[3]{cd}\\=\ \textgreater \ GCD(x,y)=3.$

Product  c*d must be a cube of some integer.   In general  c*d can be any integer, like 1 or 2,  we choose  GCD = 3.