Explain in detail why a sum of odd powers is factorable, but why a sum of even powers is not reducible to linear factors.

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n=1

`x^1+y^1=(x+y-y)^1+y^1`

`=(x+y)^1-y^1+y^1`

`=(x+y)`

When n=1, expression is linear.

n=2

`x^2+y^2=(x+y-y)^2+y^2`

`=(x+y)^2-2y(x+y)+(-y)^2+y^2`

`=(x+y)^2-2y(x+y)+2y^2`

On RHS there is no factor is common. If `(-y)^2=-y^2` then we can get common factor but `(-y)^2!=-y^2.`

Consider n=3

`x^3+y^3= (x+y-y)^3+y^3`

`=(x+y)^3-3y(x+y-y)(x+y)+(-y)^3+y^3`

`=(x+y)^3-3xy(x+y)-y^3+y^3`

`=(x+y)((x+y)^2-3xy)`

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.

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.` `

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