Yes: $$x=\frac{\left(a+b \sqrt{c}\right)^n+\left(a-b \sqrt{c}\right)^n}{2}$$ and $$y=\frac{\left(a+b \sqrt{c}\right)^n-\left(a-b \sqrt{c}\right)^n}{2 \sqrt{c}}.$$
In particular, letting $a=b=c=1$, you attain your stated goal of making "summing even/odd binomial coefficients easier": $$\sum_{k=0}^{\lfloor n/2\rfloor} \binom{n}{2k} =\sum_{k=0}^{\lfloor (n-1)/2\rfloor} \binom{n}{2k+1}=2^n/2.$$