Before I obtain flamed, correct this is a homework question, although its not being significant as I"m discovering this food on the side, and will take it it in september.The concerns is:Show the (x - y) is a element of xn - yn. Ns don"t have actually a reservation how...Thanks
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Join date Jul 1999Location Huntingdon Valley, PA 19006Posts 1,151
(x - y)*(xn-1 + xn-2*y + . . .

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+ yn-1) = xn - ynFor example, the following.(x - y)*(x + y) = x2 - y2(x - y)*(x2 + x*y + y2) = x3 - y3(x - y)*(x3 + x2*y + x*y2 + y3) = x4 - y4Sorry if there are any type of typo"s in the above.
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what Guv stated was true, but it does no prove the x-y is a factor of x^n-y^nyou can use the aspect theorem: if x-a is a aspect of polynomial f(x), then f(a)=0f(x):=x^n-y^nand we view that f(y)=0 ie x-y is a factor of f(x)=x^n-y^n
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very cute.I to be considering posting multiplied out versions the the following.x*(xn-1 + xn-2*y + . . . + yn-1)-y*(xn-1 + xn-2*y + . . . + yn-1)It is basic to display that every point cancels out yet the following.

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mods occupational nicely too:x == y (mod ) ->easy to see, just take y from every side=> xn - yn (mod )== yn - yn (mod )== 0 (mod )=> is a variable of xn - yn P.S, i know that mods room only supposed to be because that integers, but all of the reasonable used right here works because that reals too.
Originally post by sql_lall mods occupational nicely too:x == y (mod ) ->easy to see, just take y from each side=> xn - yn (mod )== yn - yn (mod )== 0 (mod )=> is a element of xn - yn P.S, i know that mods room only claimed to be for integers, but all of the logic used right here works for reals too.

but the course...now that ns think the it, due to the fact that we room talking around factors, i guess that indicates integers...oh, and possibly you can use induction:let p(n) = xn - yn(x-y) is a aspect of p(1) p(k)-p(k-1)= xn - yn - xn-1 + yn-1 = (x-y) (something else)now, all u gotta execute is uncover the "something else"
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