Factoring the expression, we have:
[tex]\begin{gathered} 8^{2x}-27^{2y} \\ (8^x)^2-(27^y)^{2y}\text{ (Rewriting the expression as a difference of squares)} \\ (8^x+27^y)(8^x-27^y)\text{ (Factoring as a as a difference of squares}) \\ ((2^3)^x+(3^3)^y)((2^3)^x-(3^3)^y)\text{ (Using prime factorization to modify the expression)} \end{gathered}[/tex]
[tex]\begin{gathered} ((2^x)^3+(3^y)^3)((2^x)^3-(3^y)^3)(\text{ Rewriting the expressions as one sum and one difference of cubes)} \\ (2^x+3^y)((2^x)^2-(2^x)(3^y)+(3^y)^2)((2^x-3^y)((2^x)^2+(2^x)(3^y)+(3^y)^2) \\ \text{ Now we have four factors after factoring the sum and the difference of cubes.} \end{gathered}[/tex]
The final answer is:
[tex](2^x+3^y)(4^x^{}^{}-(2^x)(3^y)+9^y^{})((2^x-3^y)(4^x^{}+(2^x)(3^y)+9^y^{})[/tex]
