Injectivity and norm function on finite fields 
Let $q$ be an odd prime power. Consider the map $f:\Bbb F_{q^3} \rightarrow  \Bbb F_{q^3}$, defined by 
  $$f(x)=\alpha x^q+\alpha^q x$$
  for some fixed $\alpha \in \Bbb F_{q^3} \setminus \{ 0 \}$.
  Show that $f$ is a bijection.

Hint: If $\beta \in \ker(f)  \setminus \{ 0 \}$ consider the relative norm map $N_{\Bbb F_{q^3}/{\Bbb F_q}}(\alpha \beta^q)$.
 A: Consider
$$g(x) =f(x)-f(a)=\alpha(x-a)^q+\alpha^q(x-a)$$
since $q\equiv 0$ in our field and the binomial theorem holds. Now if $f$ has a double value, say $f(a)$ which is taken on twice, then $g$ has a two zeros. However, if $b\ne a$ is such a pair of zeros, we have
$$\alpha(b-a)^q+\alpha^q(b-a)=0\iff (b-a)^{q-1}=-(\alpha)^{q-1}.$$
Then

$$\left({b-a\over \alpha}\right)^{q-1}=-1$$

So that ${b-a\over\alpha}\not\in\Bbb F_{q}$, as elements of the base field are totally determined by the fact that they are roots of $x^q-x$. However, this implies
$$\left({b-a\over \alpha}\right)^{q^2}={b-a\over\alpha}$$
i.e. ${b-a\over\alpha}\in\Bbb F_{q^2}$ which is impossible since that field is not a sub-extension of $\Bbb F_{q^3}$. Hence no such $b$ exists.
A: I like Adam's solution a lot. I arrived at the scene late, so I'm just adding this as my best guess as to what the hint means.
The mapping $f$ is linear over the subfield $\Bbb{F}_q$, so it suffices to show that its kernel is trivial. Assume that there exists a $\beta$ such that $f(\beta)=0$. This implies that
$$
\alpha\beta^q=-\alpha^q\beta.\qquad(*)
$$
Let's appply the Frobenius automorphism to both sides:
$$
\alpha^q\beta^{q^2}=-\alpha^{q^2}\beta^q.
$$
Repeat the dose remembering that raising to power $q^3$ is the identity mapping:
$$
\alpha^{q^2}\beta=-\alpha\beta^{q^2}.
$$
Let's multiply these three equations together, and arrive at
$$
N(\alpha)N(\beta)=-N(\alpha)N(\beta).
$$
Here $N(x)=x\cdot x^q\cdot x^{q^2}$ is the relative norm map.
Because $-1\neq1$ (here we need the assumption that $q$ is odd) this implies
that $N(\alpha)N(\beta)=0$. But the norm vanishes only at zero, so we can conclude that $\alpha=0$ or $\beta=0$. The former possibility was assumed not to hold, so $\beta=0$. The claim follows.
Using Adam's idea we could deduce (assuming $\alpha\beta\neq0$) from (*) directly that
$$
\left(\frac\beta\alpha\right)^{q-1}=-1.
$$
This leads to a contradiction by Adam's argument.
