Show that $x^3-a$ is irreducible in $\mathbb{Z}_7$ unless $a=0$ or $\pm1$ Show that $x^3-a$ is irreducible in $\mathbb{Z}_7$ unless $a=0$ or $\pm1$
My Idea:
Suppose $x^3-a=(Ax+b)(Bx^2+cx+d).$ 
Then $A=B=1$ or $A=B=-1$ WLOG $A=B=1.$
Then $x^3-a=x^3+x^2(b+c)+x(bc+d)+bd\\
\Rightarrow c+b=0,d+bc=0,bd=-a.$
I can't go further from here. Help...Thank You!

It's a cubic, so you need not work so hard. Indeed, you can just check if it has a root. 
By exhaustion: $x^3-a\cong 0\implies x^3\cong a$, so
$$
0^3=0=a\\
1^3=1=a\\
2^3=1=a\\
3^3=-1=a\\
(-3)^3=1=a\\
(-2)^3=-1=a\\
(-1)^3=-1=a
$$
where equality is in the the mod $7$ sense.
