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!
Proving an algebraic identity using the axioms of field
1
$\begingroup$
field-theory
-
1You'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 $ab^2$). – 2012-03-08
-
0I 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? – 2012-03-08
-
2You 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", $a\sum b_i = \sum ab_i$. – 2012-03-08
1 Answers
1
Indeed you need distributive, associative and commutative laws to prove your statement.
In fact: $$\begin{split} (a-b)(a^2+ab+b^2) &= (a+(-b))a^2 +(a+(-b))ab+(a+(-b))b^2\\ &= a^3 +(- b)a^2+a^2b+(-b)(ab)+ab^2+(-b)b^2\\ &= a^3 - ba^2+a^2b - b(ab) +ab^2-b^3\\ &= a^3 - a^2b+a^2b - (ba)b +ab^2-b^3\\ &= a^3 -(ab)b +ab^2-b^3\\ &= a^3 -ab^2 +ab^2-b^3\\ &= a^3-b^3\; . \end{split}$$