for instance,

a(aa+ab+bb)-b(aa+ab+bb)= (a-b)(aa+ab+bb)= a^3-b^3You know better than anyone George that an instance does not make a proof.

I'm not sure how you would algebraically solve your formula, would you mind showing me the steps?

Sure, I shall illustrate Post #6 in detail.

Typically, we wanna know if a[sup]n[/sup]-b[sup]n[/sup] could be expressed as (a-b)A, where A is some polynomial.

A could be find out-

Hence

and

Using notation, the proof would be:

Hence

t and k are indexes representing integars, so we can equate them when we do the following algebra.

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for instance,

a(aa+ab+bb)-b(aa+ab+bb)= (a-b)(aa+ab+bb)= a^3-b^3

You know better than anyone George that an instance does not make a proof.

I'm not sure how you would algebraically solve your formula, would you mind showing me the steps?

]]>a(aa+ab+bb)-b(aa+ab+bb)= (a-b)(aa+ab+bb)= a^3-b^3]]>

will do

]]>Should be a simple proof by induction.

]]>Thanks mathsyperson.]]>

Presumably a ≠b, but do they all have to be integers as well, or all rationals, or something?

]]>Cany anyone of you, using mathematical induction or otherwise, give a flawless proof of (a^n-b^n) always being divisble by (a-b), if the statement is true?

I shall be glad even if someone gives a counter-proof.]]>