Binomials I was verifying that [tex]\\x^2-y^2=(x-y)(x+y) \\x^3-y^3=(x-y)(x^2+xy+y^2)[/tex] and I realized that can there is a formulation more general like the theorem binomial... my question is: exist a general theorem for sum or difference of terms^n ?
You can always proceed with the division of x^n-y^n by x-y. Then you can guess and then demonstrate what the general solution is. For example: (x^11 - y^11)/ (x - y) = x^10 + x^9 y + x^8 y^2 + x^7 y^3 + x^6 y^4 + x^5 y^5 + x^4 y^6 + x^3 y^7 + x^2 y^8 + x y^9 + y^10 Try to be specific and by considering multiple examples, you can often find the path to a generalization. Never try to be general too early.
(x^{n} - y^{n}) = (x - y)(x^{n-1} + x^{n-2}y + x^{n-3}y^{2} + ... + y^{n-1}) is considered fairly elementary, but often useful. It is fairly easy to see it is true if you just multiply the x of the first bracket by the second bracket on one line and -y from the first bracket by the second bracket on the second line you will see. A connection you should not fail to observe is that this gives you the answer to getting the sum of a geometric series which is 1 + x + x^{2} + x^{n-1} (I have made the final term x^{n} for easy comparison, but you you'll be able to see what the sum is if the final term is x^{n}). The most useful of all applications of this is when x < 1 and n is infinite.
There is a well-known formula for x^{3} + y^{3} (= (x + y)(x^{2} - xy + y^{2}), and formulas for higher odd powers of x^{n} + y^{n} are fairly well known. Hint: one factor is x + y. You can get the other factor by long division. If there's a formula for x^{n} + y^{n} + z^{n} I'm not aware of it.