Find all the roots and their multiplicities of $f(x) = x^3-1$ in the field $\Bbb Z/3$

Question

I'm not sure where to start on this one. I'm thinking maybe factor it mod $3$?

Answer 1

$x^3-1\equiv (x-1)^3\mod 3$ so you only have one root, namely $1$, and its multiplicity is manifestly $3$.

Answer 2

The field $\mathbb{Z}_3$ is a very small field. Just test the three elements $\{0,1,2\}$ to see which are roots of $f$. Only the element $1$ is a root. So you can factor out $(x-1)$ from $f$. You can continue this procedure on $f/(x-1)$, and so on, and soon you'll see that there was a much easier way to approach this (Adam Hughes' answer).

Answer 3

In general in $F_p$ we have $(x+y)^p=x^p+y^p$ then $x^3-1=x^3-1^3=(x-1)^3$