Application of Fermat's little theorem I want to show this: $p$ prime, $p \equiv 2 \bmod 3 \implies  t^{3}-a$ reducible in $\mathbf{F}_p[t]$ for all $a\in \mathbf{F}_{p}$. By using Fermat's little theorem, but I don't know how.
If somebody would show me, I'd be very glad. 
 A: We will prove a more general statement: 

If $p$ is a prime and $n$ is relatively prime to $p-1$, then the map $x \mapsto x^{n}$ is a bijection from $\mathbf F_p$ to itself. In other words, every element has a unique $n^{\text{th}}$ root in $\mathbf F_p$. 

As a simple corollary, every polynomial of the form $t^n - a$ is reducible, since it is divisible by $t-b$ where $b$ is a $n^{\text{th}}$ root of $a$. 
Now, even though the above lemma is stated for the field $\mathbf F_p$, it is convenient and more enlightening to pass to the multiplicative group $\mathbf F_p^{\times}$: the additive structure is  irrelevant and the zero element is best regarded separately here. As an exercise, verify that our first lemma follows as a consequence of the following. 

If $G$ is a finite group and $n$ is relatively prime to $|G|$, then the homomorphism $\varphi : G \to G : x \mapsto x^{n}$ is an automorphism of $G$. That is, every element in $G$ has a unique $n^{\text{th}}$ root. 

Proof. Since $G$ is finite, establishing one of injectivity and surjectivity suffices,   but we show both for fun. We need the following two facts:


*

*$a^{|G|} = 1_G$ for all $a \in G$.

*Since $\gcd(n, |G|)= 1$, by Bézout's lemma, there exist integers $u, v$ such that $u n + v|G| = 1$. 
Armed with these facts, we are ready to show


*

*Surjectivity: Given any $a \in G$, 
$$
\left( a^{u} \right)^n = a^{un} = a^{un} \cdot a^{v|G|} = a^{un + v |G|} = a,
$$
so every element of $G$ is in the image of $\varphi$.

*Injectivity: Since $\varphi$ is a homomorphism, it suffices to check that its kernel is trivial. Let $a \in G$ such that $a^n = 1_G$. Then
$$
a = a^{u n + v|G|} = \left(a^{n} \right)^u \cdot a^{v|G|} = 1_G \cdot 1_G = 1_G, 
$$
and hence we are done. 
A: You need to show that $t^3 - a$ has a root in $\mathbf F_p$. By little Fermat we know that $a^p = a$. As $p \equiv -1\ \bmod 3$, we also know that $3$ divides $p + 1$. You can put these two facts together to find a cube root $b$ of $a^2$. Now, what is the cube of $a/b$?
[Maybe it's better to think about it this way: $\mathbf F_p^*$ is a cyclic group of order $p - 1$, and this order is coprime to $3$. Once again, thanks to Srivatsan for correcting an awful blunder of mine in the original version.]
