Here's another way to see it:
Consider `x^n+y^n` as a polynomial in the variable `x.` If `x^n+y^n` has a linear factor, then it also has a zero. Conversely, any zero corresponds to a linear factor. But if `n` is even, then `x^n+y^n` is always positive for all `x` (it can only be zero when both `x` and `y` are zero, but any factorization must apply to all choices of `y,` not just the special case `y=0)` , so it can't have a real zero, and thus can't have a linear factor.
If `n` is odd, then we see that if `x=-y,` then `x^n+y^n=(-y)^n+y^n=0,` so the polynomial `x^n+y^n`must have the linear factor `x+y.` The full factorization can be found by polynomial long division.` `
When n=1, expression is linear.
On RHS there is no factor is common. If `(-y)^2=-y^2` then we can get common factor but `(-y)^2!=-y^2.`
So, we have an obeservation
(i) When power is even i.e. `x^(2m)=(-x)^(2m)` , no common factor. So we can not reduce as linear factor.
(ii)When power is odd i.e. `x^(2m+1)!=(x)^(2m+1)`, so we have common factor. Thus we can reduce as linear factor.