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I am trying to prove (based on the axioms of field) that $$a^3-b^3=(a-b)(a^2+ab+b^2)$$ So, my first thought was to use the distributive law to show that $$(a-b)(a^2+ab+b^2)=(a-b)\cdot a^2+(a-b)\cdot ab+(a-b)\cdot b^2$$ And then continuing from this point. My problem is that I'm not sure if the distributive law is enough to prove this identity. Any ideas? Thanks!

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You'll need to prove that $(-x)y = x(-y) = -(xy)$ for all $x,y$, and you'll also be using associativity and commutativity (e.g., to get from $b\cdot ab$ to $ab^2$). –  Arturo Magidin Mar 8 '12 at 21:10
    
I think I know how to do the rest, my question was if this identity (the second one I wrote) could be justified relaying on the distributive law alone? –  yotamoo Mar 8 '12 at 21:14
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You are using distributivity and associativity; the distributivity axiom only tells you what to do when you have $x(y+z)$; here you have $x(y+z+w)$, so you must view it as $x((y+z)+w) = x(y+z) + xw = (xy+xz) + xw = xy+xz+xw$ (or as $x(y+(z+w)) = xy+x(z+w)= xy+(xz+xw) = xy+xz+xw$. Or form a "generalized distributivity", $a\sum b_i = \sum ab_i$. –  Arturo Magidin Mar 8 '12 at 21:15

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Indeed you need distributive, associative and commutative laws to prove your statement.

In fact: $$\begin{split} (a-b)(a^2+ab+b^2) &= (a+(-b))a^2 +(a+(-b))ab+(a+(-b))b^2\\ &= a^3 +(- b)a^2+a^2b+(-b)(ab)+ab^2+(-b)b^2\\ &= a^3 - ba^2+a^2b - b(ab) +ab^2-b^3\\ &= a^3 - a^2b+a^2b - (ba)b +ab^2-b^3\\ &= a^3 -(ab)b +ab^2-b^3\\ &= a^3 -ab^2 +ab^2-b^3\\ &= a^3-b^3\; . \end{split}$$

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