Prove that $f=x^6+ax+5$ is reducible over $\mathbb{Z_7},\forall a\in\mathbb{Z_7}$ We have $f=x^6+ax+5\in\mathbb{Z_7}$ and we have to show that it is reducible on $\mathbb{Z_7}$, $\forall a\in\mathbb{Z_7}$. Here are all my steps:

For $a=0$ we'll get $f=x^6+5\in\mathbb{Z_7}$. But $\forall x\in\mathbb{Z_7}$ with $x\neq0\Rightarrow x^6=1$. Therefore $f\neq0$ and it means that $f$ doesn't have linear factors. How  can  we continue from here?
For $a\neq0$ and $x\neq0$ we'll get $f=ax+6$, $\forall x\in\mathbb{Z_7}$. Here the only solution for which $f$ is reducible over $\mathbb{Z_7}$ is for $x=a^{-1}$. But if $x\neq a^{-1}$ then $ f$ is not reducible over $\mathbb{Z_7}$.


How can I show that $f$ is reducible over $\mathbb{Z_7},\forall a\in\mathbb{Z_7}$ ? Where am I wrong?

 A: As you said, if $a \neq 0$ we have $f(a^{-1})=0$ which means that $f$ has a root and thus is reducible [$x-a^{-1}$ is a factor].
If $a=0$ you need to factor $x^6+5$. To do this note that $5 \equiv -100 \pmod{7}$. 
A: Note that $f(a^{-1})=0$ as you noted before when $a$ is not $0$. Then we know that $(x-a^{-1})$ is a factor. We can do a "complete the sextic" approach on the function as follows:
$f(x) = (x-a^{-1})(x^5) = x^6-a^{-1}x^5$
$f(x) = (x-a^{-1})(x^5+a^{-1}x^4) = x^6 + a^{-2}x^4$
Verify that $f(x) = (x-a^{-1})(x^5+a^{-1}x^4+a^{-2}x^3+a^{-3}x^2 + a^{-4}x + 2a) = x^6-a^{-5}x + 2ax - 2= x^6-ax + 2ax +5 = x^6+ax+5$
The last $2a$ came from the observation that $a^{-1}$ was a root of both $x^6-ax$ and $x^6+ax+5$, meaning that the two functions were some multiple of $(x-a^{-1})$ from each other.
Also, when $a=0$, we need to factorize $f(x) = x^6+5 = x^6-9 = (x^3+3)(x^3-3)$.
A: You have solved the $a\neq 0$ case with the exception of saying that $x=a^{-1}$ is the solution and no more need be said about this case; taking only the $a=0$ case, we immediately have $f(x)=x^6+5\equiv 6\pmod 7$ for all $x$, and therefore there are no linear factors in this case.  However, we do have that $5=14-9=-9+2\cdot 7\equiv -9 \pmod 7$.  This brings us to
$$f(x)=x^6+5\equiv x^6-9\pmod 7$$
$x^6-9$ is a difference of squares as $(x^3)^2$ and $3^2$, leading to the factorization $x^6-9=(x^3+3)(x^3-3)$.  The fact that there are no linear factors suggests that $3,-3$ are both cubic non-residues modulo $7$, and it is easy to verify that the only cubic residues modulo $7$ are $\{-1,1\}$.
