Easiest proof :- consider a polynomial p(x) = x^n - y^n p(y) = y^n - y^n = 0 => y is zero of polynomial p(x) therefore (x-y) is a factor of p(x) hence (x-y) is a factor of (x^n - y^n)
@@Enthalpy-- The issue is that it may not be (x - y) that is the factor for higher polynomials. It could be that the hypothetically unfactorable quadratic (x^2 - y^2) or a higher term is a multiple root that is the factor that leads to the zero when x^n is y^n.
I realize it is trivial to prove this is not the case, or rather that if it was, then there is a contradiction because it could be proved that whatever hypothetical unfactorable multiple root must in fact be factorable , and thus I really like your easiest proof. But it's not quite so easy.
The trickery of adding a zero in the middle so that you can factor. 😂😂😂
Hehe yes
Alternative proof: Use long division. x^n - y^n = (x - y) P(x, y), where P(x, y) = x^(n-1)+x^(n-2)y+...+xy^(n-2)+y^(n-1).
Easiest proof :-
consider a polynomial p(x) = x^n - y^n
p(y) = y^n - y^n = 0
=> y is zero of polynomial p(x)
therefore (x-y) is a factor of p(x)
hence (x-y) is a factor of (x^n - y^n)
@@Enthalpy-- The issue is that it may not be (x - y) that is the factor for higher polynomials. It could be that the hypothetically unfactorable quadratic (x^2 - y^2) or a higher term is a multiple root that is the factor that leads to the zero when x^n is y^n.
I realize it is trivial to prove this is not the case, or rather that if it was, then there is a contradiction because it could be proved that whatever hypothetical unfactorable multiple root must in fact be factorable , and thus I really like your easiest proof. But it's not quite so easy.
To prove x-y as a factor just substitute x=y to the eq we get y^n-y^n =0
Genius 👍
Awesome!