Let $F$ be a field of characteristic 2 with more than 2 elements. Show that there are elements $a$ and $b$ in $F$ such that $(a+b)^3 \not= a^3 + b^3$.
$F$ couldn't possibly have less than 2 elements, and if it had exactly 2 — that is, $F = \mathbb Z_2$ —, $(a+b)^3$ would actually always be $a^3+b^3$. With $\#F>2$, how do I find $a$ and $b$ to violate that?
All I could do so far is simplify the inequation, using the characteristic and the commutative prroperty:
$\begin{align} (a+b)^3 &\not= a^3 + b^3\\ a^3 + 3a^2b + 3ab^2 + b^3 &\not= a^3 + b^3\\ a^3 + a^2b + ab^2 + b^3 &\not= a^3 + b^3\\ a^2b + ab^2 &\not= 0\\ ab(a + b) &\not= 0\quad. \end{align}$