Show that $x^3-3$ is irreducible in $\Bbb Z_7[x]$. 
Show that $x^3-3$ is irreducible in $\Bbb Z_7[x]$.

In the text, we haven't gotten to the theorem that the roots of polynomials are the only factors , and I would rather not prove it in this question. So, I will not be able to say that $$(x-3^{1/3})(x^2+3^{1/3}x+3^{2/3})$$ are the only factors, which do not lie in $\Bbb Z_7[x]$
 A: If $x^3-3$ is reducible, then it have a root: $x^3-3=(x-\alpha)^3$ or $x^3-3=(x-\alpha)(x^2+\beta x + \gamma)$ ($3 = 1 + 1 + 1$ or $3 = 1 + 2$, ha-ha). But $x^3-3$ haven't a root in $\mathbb Z_7$
A: To show a cubic polynomial is reducible one has to show it factorizes as a product of linear polynomial and a quadratic polynomial or a product of 3 linear polynomials. I'll describe the former case.
By multiplying the quadratic by a suitable $a$ and the linear polynomial by $a^{-1}$ we can assume the factorization is of the form:
$$ x^3-3 = (ax+b)(x^2+cx+d)$$ Now comparing coefficients you only need to solve a linear system (for the unknowns $a,b,c,d$) and prove that no solution exists.
A: If a cubic polynomial is reducible over a field, then one of its factors must have degree $1$ and so the polynomial has a root in that field.
The cubes in $\Bbb Z_7$ are $0$ and $\pm 1$ because $a=b^3$ with $a\ne0$ implies $a^2=b^6=1$ and so $a=\pm1$.
So $3$ is not a cube and so $x^3-3$ is irreducible in$\Bbb Z_7[x]$.
A: There are just $\frac{7-1}{3}=2$ cubic residues in $\mathbb{F}_7^*$, and they are obviously $\pm 1$. 
Since a third degree polynomial is irreducible over a finite field iff it has no root in the field,
$$ x^3-3 $$
is irreducible over $\mathbb{F}_7$, because $3\neq \pm 1$.
