how to separate an equation (if possible) in a desired form

Question

I want to write an equation in such a manner that (if possible) it is written as some variable function of desired variables. For example to check whether a statistic is a sufficient statistic or not we check the interaction of a parameter with the data only through the function of data.

For instance, in the following, I could write the R.H.S as $\lambda$ interacting only with $(x+y).$ Is it possible to do that in Mathematica?

$\left(e^{\lambda } (p-1) (c \lambda )^x-p e^{c \lambda } \lambda ^x\right)\left(e^{\lambda } (p-1) (c \lambda )^y-p e^{c \lambda } \lambda ^y\right) \\ = e^{2c\lambda}p^2\lambda^{(x+y)}+e^{-2c\lambda+\lambda}(1-p)p c^{y}\lambda^{(x+y)}+ e^{\lambda-c\lambda}(1-p)c^xp\lambda^{(x+y)}+e^{2\lambda}(1-p)^{2}c^{(x+y)}\lambda^{(x+y)} $

(-E^(c \[Lambda]) p \[Lambda]^x + 
   E^\[Lambda] (-1 + p) (c \[Lambda])^
    x) (-E^(c \[Lambda]) p \[Lambda]^y + 
   E^\[Lambda] (-1 + p) (c \[Lambda])^y)

Answer 1

λ interacting only with (x+y)

Is this what you mean?

eq=(-E^(c λ) p λ^x+E^λ (-1+p) (c λ)^x) (-E^(c λ) p λ^y+E^λ (-1+p) (c λ)^y)
Simplify[Collect[PowerExpand[eq],λ^(x+y)]]

$$ \lambda ^{x+y} \left(e^{\lambda } (p-1) c^x-p e^{c \lambda }\right) \left(e^{\lambda } (p-1) c^y-p e^{c \lambda }\right) $$