Answer:
[tex]\large\boxed{\dfrac{(2x-3y)^2}{(3y-2x)^2}=1}[/tex]
Step-by-step explanation:
[tex]\dfrac{(2x-3y)^2}{(3y-2x)^2}=\dfrac{(2x-3y)^2}{\left(-1(2x-3y)\right)^2}\qquad\text{use}\ (ab)^n=a^nb^n\\\\=\dfrac{(2x-3y)^2}{(-1)^2(2x-3y)^2}=\dfrac{(2x-3y)^2}{1(2x-3y)^2}=\dfrac{(2x-3y)^2}{(2x-3y)^2}=1[/tex]
