What sorcery is this?

For a quick solution, one may want to remove choices of $(x\pm y)^{2n}$ as if these answers are correct, the binomial coefficient shall be $\binom{2n}{j}$ for natural number $j < 2n$ , and the coefficient of the first term shall be a multiple of $2$ , both equations of which failed to satisfy. Now we try to expand $(x+y)^n$ and $(x-y)^n$ which gives $(x+y)^n = \sum_{i=0}^{n}\binom{n}{i}x^{n-i}y^{i}$ and $(x-y)^n = \sum_{i=0}^{n}\binom{n}{i}x^{n-i}(-y)^{i}$ At this form, if we add the two equation together, every even-degree term of $x$ or $y$ ( NOT where both $x$ and $y$ are in odd-degree) will be combined to twice its term while odd-degree term (of $x$ or $y$ ) will be subtracted to zero (it will be another way round if the two equation were subtracted), hence, the answer should be $(x+y)^n+(x-y)^n$