Help with simplifying this expression containing binomials/indices?

Question

I'm currently working on Neyman Pearson in Statistical Inference. The solution to this particular question has that:

$$ [(3CX)(1/4)^{x}(3/4)^{3-x}]/[(3CX)(3/4)^x(1/4)^{3-x}] $$

gives us:

$$ [3^{3-x}]/[3^x] $$

I'm at a loss as to how we get this result. I've tried to work it out by hand and I am assuming that the two $ (3CX)'s $ cancel. Still not sure how we end up with a $ 3^{something} $ on top and bottom. If anyone could point me in the right direction it would be greatly appreciated. (Apologies for a rather loose handed application of LaTeX, coding is not my strong point)

Answer 1

$(1/4)^x/(3/4)^x$ leaves $(1/3)^x$ and $(3/4)^{3-x}/(1/4)^{3-x}$ leaves $3^{3-x}.$

Answer 2

$$ \frac{(3CX)(1/4)^{x}(3/4)^{3-x}}{(3CX)(3/4)^x(1/4)^{3-x}}=\frac{(1/4)^{x}(3/4)^{3-x}}{(3/4)^x(1/4)^{3-x}} =(1/4)^{2x-3}(3/4)^{3-2x} =(1/4)^{2x-3}(1/4)^{-(2x-3)}3^{3-2x} =3^{3-2x} =3^{3-x}/3^x. $$