<s>[INST] How to prove that given a binary operator $*$, $a * b = a^2 - ab + b^2$ is associative? No matter how hard I try I cannot seem to prove that, given the binary operator $*$, the operation $a * b = a^2 - ab + b^2$ is associative.
This is what I have tried:
$(a * b) * c = a * (b * c) [Associativity];
$(a^2 - ab + b^2) * c = a * (b^2 - bc + c^2)$;
I have tried using $(a - b)^2$: $(a - b)^2 * c = a * (b - c)^2$;
Like other associativity problems, I could prove that but reaching an equality after expanding both sides, but that doesn't seems to be the case here. What am I doing wrong? Is this even an associative operation?
 [/INST] Just do it:
$(a*b)*c = (a^2 - ab + b^2)*c = (a^2 - ab + b^2)^2-a^2c +abc-b^2c+c^2=a^4 -2a^3b + 3a^2b^2 -2ab^3+b^4-a^2c+abc-b^2c+c^2$
$a*(b*c) = a^2-a (b^2-bc+c^2) +(b^2 - bc + c^2)^2$
It's pretty clear these need not be equal.  One is "dominated" by $a^4$ and $b^4$ while the other is dominated by $c^4$ and $b^4$.
Let $a=100$, $b=c=0$.  Then $(a*b)*c = (a*0)^2 -(a*0)0+0^2=(a*0)^2=(100^2-0*100+0)^2=100^4$.  Whereas $a*(b*c)=100^2 -100 (0*0)+(0*)^2=100^2-100 (0^2-0*0+0^2)+(0^2-0*0+0^2)^2=100^2$
</s>