I need to reduce $x^3+6=7$ over $\mathbb{Z}_7$. My approach is to factor the polynomial into three irreducible (linear) factors.
This polynomial has at most three roots (up to the equivalence classes). So one can find these roots in order to then factor it into three factors of degree one.
But here's the problem: how does one solve the equation $x^3+6=7$ $\mod7$ other than by guessing?
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The initial strategy I would use is to make the expression on the left hand side of $$ x^3+6=7 \hspace{4mm}\mbox{ in }\mathbb{Z}_7 $$ into a cyclotomic polynomial form, i.e., write it as: $$ x^3-1 \equiv 0 \mod 7. $$ We know that $x^3-1$ factors as $(x-1)(x^2+x+1)$.
Now try to factor the remaining polynomial $\mod 7$: $$ (x-1)(x^2+x+1)=(x-1)(x^2-6x+8)=(x-1)(x-2)(x-4) \mbox{ in }\mathbb{Z}_7. $$
$$\begin{align*} x^3+6&=7\pmod{7}\\ x^3-1&=0\pmod{7}\\ (x-1)(x^2+x+1)&=0\pmod{7}\\ (x-1)(x-2)(x-4)&=0\pmod{7} \end{align*}$$
To verify this, multiply as you normally do but remember to reduce coefficients $\pmod{7}$.