I have to study the existence condition of the following algebraic fraction: $$p=\frac{1}{a^3-b^3}$$ Since I can decompose the difference of the cube of $a$ and $b$ in this way: $a^3-b^3=(a-b)(a^2+ab+b^2)$, I can rewrite $p$ as: $$p=\frac{1}{(a-b)(a^2+ab+b^2)}$$ So the existence conditions will correspond to the denominator different from $0$ and so: $\begin{cases} a\neq b\\ a^2+ab+b^2\neq 0 \end{cases}$
I have a doubt: my Professor says that the terms "$a^2+ab+b^2$" is what we can call with "false square" and it is always $>0$.
But I think it is $\geq 0$, since if $a=b=0$ then it is null. I have found out that the false square is always $\neq 0$ also on the web, so I can't understand why. Can you hlep me?
$a^2+ab+b^2=\Big(a+\frac{b}{2}\Big)^2+\frac{3b^2}{4}$ and the sum of two squares is always positive, unless $a=b=0$.